Power Series

This chapter meticulously explores power series, detailing their convergence criteria and essential analytical properties. A thorough understanding of these concepts is paramount for the CUET PG MA examination, as power series form a fundamental cornerstone of Real Analysis and are consistently featured in problem-solving scenarios.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Convergence of Power Series |
| 2 | Properties of Power Series |

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We begin with Convergence of Power Series.

Part 1: Convergence of Power Series

Power series are fundamental in analysis, offering a means to represent functions as infinite sums. Understanding their convergence properties is critical for their application in calculus, differential equations, and complex analysis, making it a recurring topic in competitive examinations. We investigate the methods for determining the domain over which such series converge.

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Core Concepts

1. Definition of a Power Series

A power series is an infinite series of the form $\sum_{n=0}^{\infty} c_n (x-a)^n$, where $c_n$ are coefficients, $x$ is a variable, and $a$ is the center of the series. For simplicity, we often consider series centered at $a=0$, i.e., $\sum_{n=0}^{\infty} c_n x^n$.

2. Radius of Convergence

For every power series, there exists a unique non-negative real number $R$, called the radius of convergence, such that the series converges absolutely for $|x-a| < R$ and diverges for $|x-a| > R$. At the endpoints, $|x-a|=R$, the series may converge or diverge.

Quick Example: Ratio Test

We determine the radius of convergence for the series $\sum_{n=1}^{\infty} \frac{n^2}{2^n} x^n$.

Step 1: Identify the coefficient $c_n$.
>

Step 2: Compute the limit $L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$.
>

Step 3: Calculate $R = \frac{1}{L}$.
>

Answer: $\boxed{2}$

:::question type="MCQ" question="The radius of convergence of the power series $\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} x^n$ is:" options=["$1/4$","$2$","$4$","$\infty$"] answer="$4$" hint="Apply the Ratio Test. Recall that

" solution="Step 1: Identify $c_n$.
>

Step 2: Compute $L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$.
>

Step 3: Calculate $R = \frac{1}{L}$.
>

Answer: $\boxed{4}$
"
:::

Quick Example: Root Test

We find the radius of convergence for the series $\sum_{n=1}^{\infty} \left(\frac{n+1}{n}\right)^n x^n$.

Step 1: Identify the coefficient $c_n$.
>

Step 2: Compute the limit $L = \lim_{n \to \infty} \sqrt[n]{|c_n|}$.
>

Step 3: Calculate $R = \frac{1}{L}$.
>

Answer: $\boxed{1}$

:::question type="MCQ" question="The radius of convergence of the power series $\sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^{n^2} z^n$ is:" options=["$e$","$1/e$","$1$","$\infty$"] answer="$e$" hint="Use the Root Test. Recall that

" solution="Step 1: Identify $c_n$.
>

Step 2: Compute $L = \lim_{n \to \infty} \sqrt[n]{|c_n|}$.
>

Step 3: Calculate $R = \frac{1}{L}$.
>

Answer: $\boxed{e}$
"
:::

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3. Interval of Convergence

The interval of convergence is the set of all $x$ values for which the power series converges. It is determined by the radius of convergence $R$ and the behavior of the series at the endpoints $x = a \pm R$.

For a series centered at $a$:
* If $R=0$, the interval is $\{a\}$.
* If $R=\infty$, the interval is $(-\infty, \infty)$.
* If $0 < R < \infty$, the series converges absolutely on $(a-R, a+R)$. We must test the endpoints $x=a-R$ and $x=a+R$ separately using appropriate convergence tests for series of constants.

Quick Example: Interval of Convergence

We find the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{x^n}{n}$.

Step 1: Find the radius of convergence using the Ratio Test.
> $c_n = \frac{1}{n}$
>

>

Step 2: Determine the interval of absolute convergence.
Since $a=0$ and $R=1$, the series converges absolutely on $(-1, 1)$.

Step 3: Test the endpoints.
* At $x=1$: The series becomes $\sum_{n=1}^{\infty} \frac{1^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}$. This is the harmonic series (a p-series with $p=1$), which diverges.
* At $x=-1$: The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$. This is an alternating series.
* $b_n = \frac{1}{n}$ is positive.
* $b_n$ is decreasing.
* $\lim_{n \to \infty} \frac{1}{n} = 0$.
By the Alternating Series Test, this series converges.

Step 4: Combine results for the interval of convergence.
The series converges on $[-1, 1)$.

Answer: The interval of convergence is $[-1, 1)$.

:::question type="MCQ" question="The series $\sum_{n=1}^{\infty} \frac{x^n}{n \cdot 2^n}$ is convergent in the interval:" options=["$[-2, 2)$","$(0, 2]$","$(0, 2)$","$(-\infty, \infty)$"] answer="$[-2, 2)$" hint="First find the radius of convergence using the Ratio Test, then check the endpoints carefully." solution="Step 1: Find the radius of convergence.
> $c_n = \frac{1}{n \cdot 2^n}$
>

>

Step 2: Determine the interval of absolute convergence.
Since $a=0$ and $R=2$, the series converges absolutely on $(-2, 2)$.

Step 3: Test the endpoints.
* At $x=2$: The series becomes $\sum_{n=1}^{\infty} \frac{2^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{1}{n}$. This is the harmonic series (a p-series with $p=1$), which diverges.
* At $x=-2$: The series becomes $\sum_{n=1}^{\infty} \frac{(-2)^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n 2^n}{n \cdot 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$. This is the alternating harmonic series, which converges by the Alternating Series Test.

Step 4: Combine results for the interval of convergence.
The series converges on $[-2, 2)$. The interval is $[-2, 2)$.
Answer: \boxed{[-2, 2)}
"
:::

4. Differentiation and Integration of Power Series

Within its interval of convergence, a power series can be differentiated and integrated term by term. The radius of convergence remains unchanged, though the interval of convergence may change at the endpoints.

Quick Example: Differentiation of Power Series

Consider the geometric series $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x|<1$. We find the power series for $\frac{1}{(1-x)^2}$.

Step 1: Recall the known power series for $\frac{1}{1-x}$.
>

Step 2: Differentiate both sides with respect to $x$.
>

>

Step 3: Adjust the index for standard form (optional, but good practice).
Let $k = n-1$, so $n=k+1$. When $n=1$, $k=0$.
>

or, using $n$ as the index again:
>

Answer: The power series for $\frac{1}{(1-x)^2}$ is $\sum_{n=0}^{\infty} (n+1) x^n$. Its radius of convergence is $R=1$, same as the original series.

:::question type="MCQ" question="The power series representation for $\ln(1-x)$ for $|x|<1$ is:" options=["$\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$","$\sum_{n=1}^{\infty} \frac{x^n}{n}$","$-\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$","$-\sum_{n=1}^{\infty} \frac{x^n}{n}$"] answer="$-\sum_{n=1}^{\infty} \frac{x^n}{n}$" hint="Consider the integral of the geometric series $\frac{1}{1-x}$." solution="Step 1: Recall the geometric series.
>

Step 2: Integrate both sides from $0$ to $x$.
>

>
>

Step 3: Express $\ln(1-x)$ and adjust index.
>

Let $k=n+1$. When $n=0$, $k=1$.
>
Using $n$ as the index again:
>
Answer: \boxed{$-\sum_{n=1}^{\infty} \frac{x^n}{n}$}
"
:::

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Advanced Applications

Combining the techniques for radius and interval of convergence allows us to analyze more complex series. We observe that careful application of convergence tests at endpoints is crucial.

Worked Example:

We determine the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(x-3)^n}{n \cdot 4^n}$.

Step 1: Find the radius of convergence.
The series is centered at $a=3$. $c_n = \frac{1}{n \cdot 4^n}$.
>

>

Step 2: Determine the interval of absolute convergence.
With $a=3$ and $R=4$, the series converges absolutely for $|x-3| < 4$.
>

>
The interval of absolute convergence is $(-1, 7)$.

Step 3: Test the endpoints.
* At $x = 3-R = 3-4 = -1$:
The series becomes $\sum_{n=1}^{\infty} \frac{(-1-3)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-4)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-1)^n 4^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$.
This is the alternating harmonic series, which converges by the Alternating Series Test.
* At $x = 3+R = 3+4 = 7$:
The series becomes $\sum_{n=1}^{\infty} \frac{(7-3)^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{4^n}{n \cdot 4^n} = \sum_{n=1}^{\infty} \frac{1}{n}$.
This is the harmonic series (p-series with $p=1$), which diverges.

Step 4: Combine results for the interval of convergence.
The series converges on $[-1, 7)$.

Answer: The interval of convergence is $[-1, 7)$.

:::question type="NAT" question="The series $\sum_{n=1}^{\infty} \frac{(x+1)^n}{n^2}$ has an interval of convergence $[A, B]$. What is the value of $A+B$?" answer="$-2$" hint="Find the radius of convergence, then check the endpoints. The interval of convergence will be closed at both ends." solution="Step 1: Find the radius of convergence.
The series is centered at $a=-1$. $c_n = \frac{1}{n^2}$.
>

>

Step 2: Determine the interval of absolute convergence.
With $a=-1$ and $R=1$, the series converges absolutely for $|x-(-1)| < 1$, i.e., $|x+1| < 1$.
>

>
The interval of absolute convergence is $(-2, 0)$.

Step 3: Test the endpoints.
* At $x = -1-R = -1-1 = -2$:
The series becomes $\sum_{n=1}^{\infty} \frac{(-2+1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$.
Consider the absolute values: $\sum_{n=1}^{\infty} \left| \frac{(-1)^n}{n^2} \right| = \sum_{n=1}^{\infty} \frac{1}{n^2}$. This is a p-series with $p=2 > 1$, which converges. Since the series of absolute values converges, the original series converges absolutely at $x=-2$.
* At $x = -1+R = -1+1 = 0$:
The series becomes $\sum_{n=1}^{\infty} \frac{(0+1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}$.
This is a p-series with $p=2 > 1$, which converges.

Step 4: Combine results for the interval of convergence.
The series converges on $[-2, 0]$.
Thus, $A = -2$ and $B = 0$.
$A+B = -2+0 = -2$.
Answer: \boxed{-2}
"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n^n}{n!} x^n$ is:" options=["$e$","$1/e$","$1$","$\infty$"] answer="$1/e$" hint="Use the Ratio Test. Recall that $\lim_{n \to \infty} (1 + 1/n)^n = e$." solution="Step 1: Identify $c_n$.
>

Step 2: Compute $L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$.
>

Step 3: Calculate $R = \frac{1}{L}$.
>

Answer: \boxed{1/e}"
:::

:::question type="MCQ" question="The interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(x-5)^n}{n \ln n}$ is:" options=["$[4, 6)$","$(4, 6]$","$(4, 6)$","$[4, 6]$"] answer="$[4, 6)$" hint="Find R using the Ratio Test, then test endpoints. For $x=4$, use the Alternating Series Test. For $x=6$, use the Integral Test or Comparison Test with a known divergent series." solution="Step 1: Find the radius of convergence.
The series is centered at $a=5$. $c_n = \frac{1}{n \ln n}$.
>

>

Step 2: Determine the interval of absolute convergence.
With $a=5$ and $R=1$, the series converges absolutely for $|x-5| < 1$.
>

>
The interval of absolute convergence is $(4, 6)$.

Step 3: Test the endpoints.
* At $x = 4$: The series becomes
>

This is an alternating series. Let $b_n = \frac{1}{n \ln n}$.
1. $b_n > 0$ for $n \ge 2$.
2. $b_n$ is decreasing for $n \ge 2$.
3. $\lim_{n \to \infty} \frac{1}{n \ln n} = 0$.
By the Alternating Series Test, the series converges at $x=4$.
* At $x = 6$: The series becomes
>
We use the Integral Test. Let $f(x) = \frac{1}{x \ln x}$. This function is positive, continuous, and decreasing for $x \ge 2$.
>
Since the integral diverges, the series diverges at $x=6$.

Step 4: Combine results for the interval of convergence.
The series converges on $[4, 6)$.
Answer: \boxed{[4, 6)}"
:::

:::question type="NAT" question="If the series $\sum_{n=0}^{\infty} \frac{(n!)^k}{(kn)!} x^n$ has a radius of convergence $R$, then for $k=2$, $R$ is:" answer="4" hint="Apply the Ratio Test. Be careful with factorials." solution="Step 1: Identify $c_n$ for $k=2$.
>

Step 2: Compute $L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$.
>

Step 3: Calculate $R = \frac{1}{L}$.
>

Answer: \boxed{4}"
:::

:::question type="MSQ" question="Which of the following statements about the power series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n} (x-1)^n$ are correct?" options=["The radius of convergence is $1$.","The series converges at $x=0$.","The series converges at $x=2$.","The interval of convergence is $(0, 2)$."] answer="The radius of convergence is $1$,The series converges at $x=2$" hint="First find R. Then test endpoints $x=0$ and $x=2$." solution="Step 1: Find the radius of convergence.
The series is centered at $a=1$. $c_n = \frac{(-1)^n}{n}$.
>

>
Thus, Statement 1: 'The radius of convergence is $1$' is correct.

Step 2: Determine the interval of absolute convergence.
With $a=1$ and $R=1$, the series converges absolutely for $|x-1| < 1$, which means $0 < x < 2$.

Step 3: Test the endpoints.
* At $x=0$: Substitute $x=0$ into the series:
>

This is the harmonic series, which diverges.
Thus, Statement 2: 'The series converges at $x=0$' is incorrect.
* At $x=2$: Substitute $x=2$ into the series:
>
This is the alternating harmonic series. It satisfies the conditions of the Alternating Series Test ($b_n = 1/n$ is positive, decreasing, and $\lim_{n \to \infty} 1/n = 0$). Therefore, this series converges.
Thus, Statement 3: 'The series converges at $x=2$' is correct.

Step 4: Determine the full interval of convergence.
The series converges for $x \in (0, 2]$ (from step 2 and endpoint tests).
Thus, Statement 4: 'The interval of convergence is $(0, 2)$' is incorrect.

Answer: \boxed{\text{The radius of convergence is } 1 \text{, The series converges at } x=2}}"
:::

:::question type="MCQ" question="The series $\sum_{n=0}^{\infty} \frac{x^{2n}}{2^n}$ has an interval of convergence:" options=["$(-2, 2)$","$(-\sqrt{2}, \sqrt{2})$","$(-\infty, \infty)$","$(-1, 1)$"] answer="$(-\sqrt{2}, \sqrt{2})$" hint="Treat $y=x^2$ and find the convergence for $y$. Or apply the Ratio Test carefully to $x^{2n}$." solution="Method 1: Using Ratio Test for $x^{2n}$
Step 1: Identify $c_n$ for the term $x^n$. Here, the series is $\sum_{n=0}^{\infty} c_n x^n$ where $c_n = 0$ for odd $n$, and $c_n = 1/2^{n/2}$ for even $n$. This makes the standard Ratio/Root Test slightly awkward.

Instead, we apply the Ratio Test to the form $\sum_{n=0}^{\infty} a_n$, where $a_n = \frac{x^{2n}}{2^n}$.
>

For convergence, we require $L' < 1$.
>
>
>
>

Step 2: Test endpoints.
* At $x = \sqrt{2}$: The series becomes
>

This series diverges.
* At $x = -\sqrt{2}$: The series becomes
>
This series diverges.

Step 3: Combine results.
The interval of convergence is $(-\sqrt{2}, \sqrt{2})$.

Method 2: Substitution
Let $y = x^2$. The series becomes $\sum_{n=0}^{\infty} \frac{y^n}{2^n} = \sum_{n=0}^{\infty} \left(\frac{y}{2}\right)^n$.
This is a geometric series with ratio $r = y/2$. It converges if $|r| < 1$.
>

>
Substitute back $y=x^2$:
>
>
>
At the endpoints $x=\pm\sqrt{2}$, $y=x^2=2$. The series becomes $\sum_{n=0}^{\infty} (2/2)^n = \sum_{n=0}^{\infty} 1$, which diverges.
The interval of convergence is $(-\sqrt{2}, \sqrt{2})$.
Answer: \boxed{(-\sqrt{2}, \sqrt{2})}"
:::

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Summary

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What's Next?

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Part 2: Properties of Power Series

Power series are fundamental in real analysis, providing a means to represent functions as infinite sums of powers. We utilize them for function approximation, solving differential equations, and evaluating integrals, making them a recurring topic in competitive examinations.

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Core Concepts

1. Definition of a Power Series

A power series is an infinite series of the form $\sum_{n=0}^{\infty} c_n (x-a)^n$, where $a$ is the center of the series, $x$ is a variable, and $c_n$ are constants known as the coefficients. When $a=0$, the series is a Maclaurin series.

Quick Example:

Consider the power series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

Step 1: Identify the center and coefficients.

> The series is centered at $a=0$.
>
> The coefficients are $c_n = \frac{1}{n!}$.

Step 2: Write out the first few terms.

>

:::question type="MCQ" question="Which of the following is a power series centered at $x=2$?" options=["$\sum_{n=0}^{\infty} n! x^n$","$\sum_{n=0}^{\infty} \frac{(x-2)^n}{n}$","$\sum_{n=0}^{\infty} \frac{x^n}{2^n}$","$\sum_{n=0}^{\infty} (x+2)^n$"] answer="$\sum_{n=0}^{\infty} \frac{(x-2)^n}{n}$" hint="A power series centered at $a$ has the form $\sum c_n (x-a)^n$." solution="For a power series centered at $a$, the term involving $x$ must be $(x-a)$. In this case, $a=2$, so we look for $(x-2)^n$. Option B matches this form."
:::

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2. Radius of Convergence (R)

The radius of convergence $R$ of a power series $\sum_{n=0}^{\infty} c_n (x-a)^n$ is a non-negative number (or $\infty$) such that the series converges for $|x-a| < R$ and diverges for $|x-a| > R$. The Ratio Test or Root Test are commonly employed to determine $R$.

Quick Example:

Determine the radius of convergence for the series $\sum_{n=1}^{\infty} \frac{(x-3)^n}{n}$.

Step 1: Identify $c_n$ and $a$.

> Here $a=3$ and $c_n = \frac{1}{n}$.

Step 2: Apply the Ratio Test.

> We calculate $\lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$.
>
>

>
> Thus, $L=1$.

Step 3: Determine $R$.

> Since $L=1$, $R = \frac{1}{L} = \frac{1}{1} = 1$.

Answer: The radius of convergence is $R=1$.

:::question type="MCQ" question="Find the radius of convergence for the power series $\sum_{n=0}^{\infty} \frac{n!}{2^n} x^n$." options=["$0$","$1$","$2$","$\infty$"] answer="$0$" hint="Use the Ratio Test. Pay attention to the limit value." solution="Step 1: Identify $c_n$.
> Here $c_n = \frac{n!}{2^n}$.
>
> Step 2: Apply the Ratio Test.
>

>
> Step 3: Determine $R$.
> Since $L=\infty$, the radius of convergence $R=0$. The series only converges at its center $x=0$."
:::

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3. Interval of Convergence

The interval of convergence is the set of all $x$ values for which the power series converges. It is centered at $a$ and has length $2R$. After finding $R$, we must check the convergence at the endpoints $x = a-R$ and $x = a+R$ separately, as the Ratio Test is inconclusive at these points.

Quick Example:

Determine the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(x-3)^n}{n}$. (From previous example, $R=1$ and $a=3$).

Step 1: Establish the open interval of convergence.

> The series converges for $|x-3| < 1$, which means $-1 < x-3 < 1$.
> Adding 3 to all parts, we get $2 < x < 4$.

Step 2: Check the left endpoint $x=a-R = 3-1=2$.

> Substitute $x=2$ into the series:
>

> This is the alternating harmonic series, which converges by the Alternating Series Test.

Step 3: Check the right endpoint $x=a+R = 3+1=4$.

> Substitute $x=4$ into the series:
>

> This is the harmonic series, which diverges.

Step 4: State the interval of convergence.

> The series converges for $x \in [2, 4)$.

Answer: The interval of convergence is $[2, 4)$.

:::question type="MCQ" question="Find the interval of convergence for the power series $\sum_{n=1}^{\infty} \frac{(x+1)^n}{n \cdot 2^n}$." options=["$(-3, 1)$","$( -3, 1 ]$","$[ -3, 1 )$","$[ -3, 1 ]$"] answer="$[-3, 1)$" hint="First find the radius of convergence using the Ratio Test, then check the endpoints. Remember to apply the appropriate convergence tests for the resulting series at the endpoints." solution="Step 1: Determine the radius of convergence.
> The series is centered at $a=-1$. Let $c_n = \frac{1}{n \cdot 2^n}$.
>

> So, $L=\frac{1}{2}$, which means $R = \frac{1}{L} = 2$.
>
> Step 2: Determine the open interval of convergence.
> The series converges for $|x - (-1)| < 2$, i.e., $|x+1| < 2$.
> This implies $-2 < x+1 < 2$, so $-3 < x < 1$.
>
> Step 3: Check the left endpoint $x = -3$.
> Substitute $x=-3$ into the series:
>
> This is the alternating harmonic series, which converges by the Alternating Series Test. So $x=-3$ is included.
>
> Step 4: Check the right endpoint $x = 1$.
> Substitute $x=1$ into the series:
>
> This is the harmonic series, which diverges. So $x=1$ is not included.
>
> Step 5: State the interval of convergence.
> Combining the results, the interval of convergence is $[-3, 1)$."
:::

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4. Properties of Power Series

Power series behave much like polynomials within their interval of convergence, allowing for term-by-term operations.

4.1. Uniqueness of Power Series

If a function $f(x)$ can be represented by a power series centered at $a$, $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$, then this representation is unique. That is, the coefficients $c_n$ are uniquely determined by $f(x)$.

4.2. Term-by-Term Differentiation

A power series can be differentiated term by term within its interval of convergence. The radius of convergence remains unchanged.

Quick Example:

Given $f(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x|<1$. Find the power series for $f'(x)$.

Step 1: Differentiate term by term.

>

Step 2: Relate to the original function's derivative.

> We know $f'(x) = \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2}$.
>
> Thus, $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}$ for $|x|<1$.

Answer: The power series for $f'(x)$ is $\sum_{n=1}^{\infty} n x^{n-1}$.

:::question type="MCQ" question="If $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for all $x$, what is the power series for $f''(x)$?" options=["$\sum_{n=0}^{\infty} \frac{x^n}{n!}$","$\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}$","$\sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!}$","$\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!}$"] answer="$\sum_{n=0}^{\infty} \frac{x^n}{n!}$" hint="Differentiate twice term by term and adjust the summation index." solution="Step 1: Differentiate $f(x)$ once.
>

> Let $k=n-1$. When $n=1$, $k=0$.
>
>
> Step 2: Differentiate $f'(x)$ (which is $f(x)$ again) once more.
>
> Let $j=k-1$. When $k=1$, $j=0$.
>
> Replacing $j$ with $n$, we get $\sum_{n=0}^{\infty} \frac{x^n}{n!}$. This is the original series.
> This matches the property that $e^x$ is its own derivative.
>
> Answer: $\boxed{\sum_{n=0}^{\infty} \frac{x^n}{n!}}$"
:::

4.3. Term-by-Term Integration

A power series can be integrated term by term within its interval of convergence. The radius of convergence remains unchanged.

Quick Example:

Given the geometric series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x|<1$. Find the power series for $\ln(1-x)$.

Step 1: Integrate the series for $\frac{1}{1-x}$.

> We know $\int \frac{1}{1-x} dx = -\ln(1-x) + C_1$.
>
> Integrating the series term by term:
>

> Let $k=n+1$. Then $n=k-1$. When $n=0$, $k=1$.
>

Step 2: Determine the constant of integration $C$.

> We have $-\ln(1-x) = C + \sum_{k=1}^{\infty} \frac{x^k}{k}$.
> Set $x=0$: $-\ln(1-0) = C + 0$, so $0 = C$.
>
> Thus, $-\ln(1-x) = \sum_{k=1}^{\infty} \frac{x^k}{k} = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$.

Step 3: Express $\ln(1-x)$.

>

> This series is valid for $|x|<1$. We must check $x=-1$.
> For $x=-1$, $\ln(2) = - \sum_{k=1}^{\infty} \frac{(-1)^k}{k} = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \dots$, which is the alternating harmonic series, convergent.
> The interval of convergence is $[-1, 1)$.

Answer: The power series for $\ln(1-x)$ is $- \sum_{n=1}^{\infty} \frac{x^n}{n}$, for $x \in [-1, 1)$.

:::question type="MCQ" question="Given the power series $\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n$ for $|x|<1$. Find the power series for $\arctan(x)$." options=["$\sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$","$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$","$\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$","$\sum_{n=0}^{\infty} \frac{x^{2n}}{2n}$"] answer="$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$" hint="Recall that $\int \frac{1}{1+x^2} dx = \arctan(x) + C$. Manipulate the given series to match the integrand." solution="Step 1: Modify the given series to represent $\frac{1}{1+x^2}$.
> Replace $x$ with $x^2$ in the series for $\frac{1}{1+x}$:
>

> This series converges for $|x^2|<1$, which means $|x|<1$.
>
> Step 2: Integrate the new series term by term.
>
>
> Step 3: Determine the constant of integration $C$.
> We know $\arctan(x) = C + \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$.
> Set $x=0$: $\arctan(0) = C + 0$, so $0=C$.
>
> Step 4: State the power series for $\arctan(x)$.
>
> The interval of convergence for $\arctan(x)$ is $[-1, 1]$.
>
> Answer: $\boxed{\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}}$"
:::

4.4. Algebra of Power Series

Power series can be added, subtracted, and multiplied like polynomials within their common interval of convergence.

Quick Example:

Find the first three non-zero terms of the power series for $e^x \sin x$ using multiplication of series.

Step 1: Write out the known Maclaurin series for $e^x$ and $\sin x$.

>

Step 2: Multiply the series term by term.

>

>
> Constant term: $0$ (since $\sin x$ has no constant term).
> Term in $x$: $1 \cdot x = x$.
> Term in $x^2$: $x \cdot x = x^2$.
> Term in $x^3$: $\frac{x^2}{2} \cdot x + 1 \cdot (-\frac{x^3}{6}) = \frac{x^3}{2} - \frac{x^3}{6} = \frac{3x^3 - x^3}{6} = \frac{2x^3}{6} = \frac{x^3}{3}$.

Answer: The first three non-zero terms are $x + x^2 + \frac{x^3}{3}$.

:::question type="MCQ" question="Given $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ and $e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}$. Find the power series for $\cosh x = \frac{e^x + e^{-x}}{2}$." options=["$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$","$\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$","$\sum_{n=0}^{\infty} \frac{x^n}{n!}$","$\sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n!}$"] answer="$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$" hint="Add the two series term by term and observe which terms cancel out or combine." solution="Step 1: Write out the series for $e^x$ and $e^{-x}$.
>

>
> Step 2: Add the two series.
>
>
> Step 3: Divide by 2 to get $\cosh x$.
>
> This can be written in summation notation as $\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$.
>
> Answer: $\boxed{\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}}$"
:::

5. Taylor and Maclaurin Series

Taylor series provide a general method to represent a function as a power series. A Maclaurin series is a special case of a Taylor series where the expansion is centered at $a=0$.

We list several common Maclaurin series, as they are frequently tested:

Quick Example:

Find the Maclaurin series for $f(x) = \sin x$.

Step 1: Find the derivatives of $f(x)$ and evaluate them at $x=0$.

> $f(x) = \sin x \implies f(0) = 0$
> $f'(x) = \cos x \implies f'(0) = 1$
> $f''(x) = -\sin x \implies f''(0) = 0$
> $f'''(x) = -\cos x \implies f'''(0) = -1$
> $f^{(4)}(x) = \sin x \implies f^{(4)}(0) = 0$
> The pattern of derivatives at $x=0$ is $0, 1, 0, -1, 0, 1, 0, -1, \dots$

Step 2: Substitute these values into the Maclaurin series formula.

>

Step 3: Write the series in summation notation.

> The terms are for odd powers of $x$ with alternating signs. The general term can be written as $(-1)^n \frac{x^{2n+1}}{(2n+1)!}$.
>
>

Answer: The Maclaurin series for $\sin x$ is $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$.

:::question type="MCQ" question="Which of the following functions has the Maclaurin series expansion $1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$?" options=["$e^x$","$e^{-x}$","$\sin x$","$\cos x$"] answer="$e^{-x}$" hint="Compare the given series with the standard Maclaurin series expansions." solution="The given series is $1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$.
This can be written as $\sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n!}$.
We know that the Maclaurin series for $e^x$ is $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.
If we replace $x$ with $-x$ in the series for $e^x$, we get:

This matches the given series.
>
> Answer: $\boxed{e^{-x}}$"
:::

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Advanced Applications

Power series can be used to approximate values, evaluate difficult integrals, or solve differential equations.

Quick Example:

Evaluate the definite integral $\int_0^1 \frac{\sin x}{x} dx$ as an infinite series.

Step 1: Write the Maclaurin series for $\sin x$.

>

Step 2: Form the series for $\frac{\sin x}{x}$.

> Divide each term by $x$ (for $x \ne 0$):
>

Step 3: Integrate the series term by term from $0$ to $1$.

>

Answer: The integral is $1 - \frac{1}{18} + \frac{1}{600} - \dots$.

:::question type="NAT" question="Using the Maclaurin series for $e^x$, approximate $e^{0.1}$ to four decimal places. (Truncate the series after the $x^3$ term)." answer="1.1052" hint="Substitute $x=0.1$ into the Maclaurin series for $e^x$ and sum the first four terms." solution="Step 1: Write the Maclaurin series for $e^x$.
>

>
> Step 2: Substitute $x=0.1$ and sum the terms up to $x^3$.
>
>
> Step 3: Calculate the sum and round to four decimal places.
>
> Rounding to four decimal places, we get $1.1052$.
>
> Answer: $\boxed{1.1052}$"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{x^n}{3^n (n+1)}$ is:" options=["$0$","$1$","$3$","$\infty$"] answer="$3$" hint="Apply the Ratio Test to find the limit of the ratio of consecutive terms." solution="Step 1: Identify $c_n$.
> Here $c_n = \frac{1}{3^n (n+1)}$.
>
> Step 2: Apply the Ratio Test.
> We calculate $L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|$.
>

>
> Step 3: Determine $R$.
> The radius of convergence $R = \frac{1}{L} = \frac{1}{1/3} = 3$.
Answer: $\boxed{3}$"
:::

:::question type="MCQ" question="The power series for $f(x) = \frac{1}{(1-x)^2}$ centered at $x=0$ is:" options=["$\sum_{n=0}^{\infty} x^n$","$\sum_{n=1}^{\infty} n x^{n-1}$","$\sum_{n=0}^{\infty} n x^n$","$\sum_{n=2}^{\infty} n(n-1) x^{n-2}$"] answer="$\sum_{n=1}^{\infty} n x^{n-1}$" hint="Recall the geometric series and its derivative." solution="Step 1: Start with the geometric series.
> We know that $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots$ for $|x|<1$.
>
> Step 2: Differentiate both sides with respect to $x$.
> The derivative of $\frac{1}{1-x}$ is $\frac{d}{dx}(1-x)^{-1} = -1(1-x)^{-2}(-1) = \frac{1}{(1-x)^2}$.
>
> Step 3: Differentiate the series term by term.
>

>
> Thus, $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}$.
Answer: $\boxed{\sum_{n=1}^{\infty} n x^{n-1}}$"
:::

:::question type="NAT" question="If $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, what is the value of $f'(0)$?" answer="1" hint="Identify the function represented by the power series and evaluate its derivative at $x=0$." solution="Step 1: Identify the function represented by the power series.
> The series $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ is the Maclaurin series for $e^x$.
> So, $f(x) = e^x$.
>
> Step 2: Find the first derivative of $f(x)$.
> $f'(x) = \frac{d}{dx}(e^x) = e^x$.
>
> Step 3: Evaluate $f'(0)$.
> $f'(0) = e^0 = 1$.
Answer: $\boxed{1}$"
:::

:::question type="MCQ" question="Consider the series $S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$. This series represents:" options=["$\ln(2)$","$\ln(1)$","$\frac{\pi}{4}$","diverges"] answer="$\ln(2)$" hint="Compare the series with the Maclaurin series for $\ln(1+x)$." solution="Step 1: Recall the Maclaurin series for $\ln(1+x)$.
>

>
> Step 2: Substitute $x=1$ into the series.
> When $x=1$, the series becomes:
>
> This exactly matches the given series $S$.
>
> The series converges at $x=1$ by the Alternating Series Test.
Answer: $\boxed{\ln(2)}$"
:::

:::question type="MSQ" question="Which of the following statements about the interval of convergence for $\sum_{n=1}^{\infty} \frac{(x-5)^n}{n^2}$ are correct?" options=["The radius of convergence is $1$.","The interval of convergence is $(4, 6)$.","The series converges at $x=4$.","The series converges at $x=6$."] answer="The radius of convergence is $1$,The series converges at $x=4$,The series converges at $x=6$" hint="Use the Ratio Test for the radius. Then check both endpoints for convergence." solution="Step 1: Find the radius of convergence.
> The series is centered at $a=5$. Let $c_n = \frac{1}{n^2}$.
>

> So, $L=1$, which means $R = \frac{1}{L} = 1$.
> Thus, 'The radius of convergence is $1$' is correct.
>
> Step 2: Determine the open interval.
> The series converges for $|x-5| < 1$, which means $-1 < x-5 < 1$, so $4 < x < 6$.
> Thus, 'The interval of convergence is $(4, 6)$' is incorrect because it doesn't account for endpoints.
>
> Step 3: Check the left endpoint $x=4$.
> Substitute $x=4$ into the series:
>
> This is an alternating series. The terms $\frac{1}{n^2}$ are positive, decreasing, and tend to 0. By the Alternating Series Test, it converges. Moreover, $\sum \left| \frac{(-1)^n}{n^2} \right| = \sum \frac{1}{n^2}$, which is a convergent p-series ($p=2 > 1$). So the series converges absolutely.
> Thus, 'The series converges at $x=4$' is correct.
>
> Step 4: Check the right endpoint $x=6$.
> Substitute $x=6$ into the series:
>
> This is a p-series with $p=2 > 1$, which converges.
> Thus, 'The series converges at $x=6$' is correct.
>
> The full interval of convergence is $[4, 6]$.
Answer: $\boxed{\text{The radius of convergence is } 1 \text{, The series converges at } x=4 \text{, The series converges at } x=6}$"
:::

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Summary

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What's Next?

Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="What is the interval of convergence for the power series $\sum_{n=1}^{\infty} \frac{(x-3)^n}{n}$?" options=["$[2, 4)$","$(2, 4]$","$(2, 4)$","$[2, 4]$"] answer="$[2, 4)$" hint="Use the Ratio Test to find the radius of convergence, then check the endpoints separately." solution="Applying the Ratio Test:

For convergence, $|x-3| < 1$, which implies $2 < x < 4$.
At $x=2$: The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, which converges by the Alternating Series Test.
At $x=4$: The series becomes $\sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series and diverges.
Thus, the interval of convergence is $[2, 4)$.
Answer: \boxed{[2, 4)}}"
:::

:::question type="NAT" question="What is the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n^n}{n!} x^n$?" answer="0.36787944" hint="Use the Ratio Test and recall the definition of $e$ as a limit." solution="Using the Ratio Test:

For convergence,

Therefore, the radius of convergence $R = 1/e \approx 0.36787944$.
Answer: \boxed{0.36787944}"
:::

:::question type="MCQ" question="Consider the function $f(x) = \sin(x^2)$. Which of the following is its Maclaurin series?" options=["$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+2}$","$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{4n}$","$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2}$","$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n+2}$"] answer="$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2}$" hint="Recall the Maclaurin series for $\sin u$ and substitute $u=x^2$." solution="The Maclaurin series for $\sin u$ is given by $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} u^{2n+1}$.
Substituting $u = x^2$, we get:

Answer: \boxed{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2}}}"
:::

:::question type="NAT" question="If $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for all $x \in \mathbb{R}$, what is $f'(1)$?" answer="2.71828" hint="Identify the function represented by the power series and then differentiate." solution="The given power series is the Maclaurin series for $e^x$, so $f(x) = e^x$.
Differentiating term-by-term:

Let $k = n-1$. Then

Therefore,

Answer: \boxed{2.71828}"
:::

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What's Next?