Sequences, Series and Limits Short Notes for GATE DA

Sequences, Series and Limits short notes for GATE DA: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Cheat Sheet: Infinite Series Core Formulas

Cheat Sheet: Core Formulas

Concept	Formula / Condition
Geometric Sum	$S = \frac{a}{1 - r}$ for $∣ r ∣ < 1$
Telescoping	$\sum (b_{n} - b_{n + 1}) = b_{1} - lim b_{N + 1}$
p-Series	$\sum \frac{1}{n ^{p}}$ converges $⟺ p > 1$
Ratio Test	$lim \frac{a _{n + 1}}{a _{n}} < 1 ⟹$ Converges
Geo. Expectation	$\sum_{n = 1}^{\infty} n p (1 - p)^{n - 1} = \frac{1}{p}$

Execution Algorithm: Double Summations

Execution Algorithm

Step-by-Step Protocol

Inspect: Look at $a_{i, j}$ . Is it separable? ( $f (i) \cdot g (j)$ )
Split: $\sum_{i} \sum_{j} f (i) g (j) ⟹ (\sum_{i} f (i)) \times (\sum_{j} g (j))$
Check Indices: Note exact starting values for $i$ and $j$ .
Evaluate 1D Sums: Apply geometric or standard series formulas.
Combine: Multiply the results.

Fallback (If NOT separable)

Evaluate the inner sum first (treat outer variable as constant).
Substitute the result into the outer sum.
Evaluate the outer sum.

Revision Cheat Sheet

Core Expansions ( $x \to 0$ )

$sin x \approx x - \frac{x ^{3}}{6}$
$cos x \approx 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$
$e^{x} \approx 1 + x + \frac{x ^{2}}{2}$
$ln (1 + x) \approx x - \frac{x ^{2}}{2}$

Shortcuts

$(1 + x)^{n} \approx 1 + n x$
$ln (A B) = ln A + ln B$

Strategy

Identify form
Choose tool
Expand & Cancel

Sequences, Series and Limits: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Calculus and Optimization MCQ

Match the double summation term $a_{i, j}$ in List I with the most appropriate evaluation method in List II.

List I:

$a_{i, j} = 2^{- i} 3^{- j}$
$a_{i, j} = \frac{1}{i + j}$

List II:

P. Evaluate inner sum first, then outer sum.

Q. Split into product of two independent single sums.

A.
1-P, 2-Q
B.
1-Q, 2-P
C.
1-Q, 2-Q
D.
1-P, 2-P

Correct Answer:

Step-by-Step Solution

Key idea: Separable terms can be split into a product of single sums; non-separable terms must be evaluated iteratively.

Step 1: Analyze term 1: $a_{i, j} = 2^{- i} 3^{- j}$ . This is separable into $f (i) g (j)$ where $f (i) = 2^{- i}$ and $g (j) = 3^{- j}$ . Thus, it can be split into the product of two independent single sums (Method Q).

Step 2: Analyze term 2: $a_{i, j} = \frac{1}{i + j}$ . The variables $i$ and $j$ are coupled in the denominator. It cannot be factored into $f (i) g (j)$ . Thus, it must be evaluated by computing the inner sum first, then the outer sum (Method P).

Step 3: Match 1 with Q, and 2 with P.

Answer: 1-Q, 2-P.

Question 2 · Calculus and Optimization MCQ

Consider the following assertion and reason regarding double summations:

Assertion (A): For a double summation $\sum_{i} \sum_{j} a_{i, j}$ with mixed positive and negative terms, swapping the order of summation always yields the same result.

Reason (R): Fubini's theorem for series requires absolute convergence to guarantee that the iterated sums are equal.

Which of the following is correct?

A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is false, but R is true.
D.
A is true, but R is false.

Correct Answer:

Step-by-Step Solution

Key idea: Swapping the order of summation is only guaranteed for absolutely convergent series.

Step 1: Analyze Assertion (A). It claims swapping always yields the same result for mixed signs. This is false. Conditionally convergent double series can yield different sums when the order is swapped.

Step 2: Analyze Reason (R). Fubini's theorem for series indeed requires $\sum_{i} \sum_{j} ∣ a_{i, j} ∣ < \infty$ (absolute convergence) to guarantee equality of iterated sums. This is true.

Step 3: Since A is false and R is true, the correct option is "A is false, but R is true."

Answer: A is false, but R is true.

Question 3 · Calculus and Optimization MCQ

For a double series $\sum_{i} \sum_{j} a_{i, j}$ , which of the following conditions makes it impossible to guarantee that $\sum_{i} \sum_{j} a_{i, j} = \sum_{j} \sum_{i} a_{i, j}$ using standard absolute convergence theorems?

A.
$a_{i, j} \geq 0$ for all $i, j$ and the sum is finite.
B.
$\sum_{i} \sum_{j} ∣ a_{i, j} ∣ = \infty$
C.
$a_{i, j} = 2^{- i} 3^{- j}$
D.
The series is separable into $f (i) g (j)$ with positive terms.

Correct Answer:

Step-by-Step Solution

Key idea: Fubini's theorem for series requires absolute convergence to guarantee the equality of iterated sums.

Step 1: Recall the condition for swapping summation order. The order can be swapped safely if the double series is absolutely convergent, i.e., $\sum_{i} \sum_{j} ∣ a_{i, j} ∣ < \infty$ .

Step 2: Analyze the options. Option A describes a positive series with a finite sum, which implies absolute convergence. Option C and D describe specific absolutely convergent positive series.

Step 3: Option B states $\sum_{i} \sum_{j} ∣ a_{i, j} ∣ = \infty$ . This means the series is not absolutely convergent. In this case, standard theorems cannot guarantee that the iterated sums are equal.

Answer: $\sum_{i} \sum_{j} ∣ a_{i, j} ∣ = \infty$ .

Question 4 · Calculus and Optimization MCQ

Match each term in List I with its handling category in List II.

List I:

P: $\frac{7 ^{- i}}{i + j}$

Q: $\frac{i}{j ^{2}}$

R: $\frac{1}{i ^{2} + j ^{2}}$

List II:

1: Fully separable as $f (i) \cdot g (j)$

2: Inner sum computable in closed form after fixing $i$

3: Neither separable nor closed-form inner evaluation

A.
P-2, Q-1, R-3
B.
P-1, Q-2, R-3
C.
P-3, Q-1, R-2
D.
P-2, Q-3, R-1

Correct Answer:

Step-by-Step Solution

Key idea: Separability needs pure multiplicative independence; otherwise check whether freezing one index yields a known inner sum.

Step 1: Q: $\frac{i}{j ^{2}} = i \cdot \frac{1}{j ^{2}}$ . That is exactly $f (i) g (j)$ with $f (i) = i$ , $g (j) = j^{- 2}$ . Category 1.

Step 2: P: $\frac{7 ^{- i}}{i + j}$ . Fix $i$ ; the inner sum over $j$ becomes $7^{- i} \sum_{j} \frac{1}{i + j}$ , a shifted harmonic tail expressible via digamma-type closed forms — at least tractable once $i$ is fixed, and clearly not separable since $i$ appears inside the denominator alongside $j$ . Category 2.

Step 3: R: $\frac{1}{i ^{2} + j ^{2}}$ . Not separable (sum inside reciprocal), and fixing $i$ leaves $\sum_{j} \frac{1}{j ^{2} + i ^{2}}$ which has no elementary closed form matching standard series tools here. Category 3.

Wrong path: declaring P separable by pulling $7^{- i}$ out and ignoring the $i$ hidden in $i + j$ leads to Option B's error.

Generalization: scan for coupling variables inside a shared denominator before claiming separability.

Verification: Q splits into two independent sums cleanly; matches category 1.

Answer: A.

Question 5 · Calculus and Optimization MCQ

Assertion (A): $lim_{x \to 0} \frac{l n (( 1 + x ^{2} ) c o s x )}{x ^{2}} = \frac{1}{2}$ .

Reason (R): The logarithm of a product splits into a sum of logarithms, allowing separate Taylor expansion of each factor near $x = 0$ .

A.
Both A and R are true, and R correctly explains A.
B.
Both A and R are true, but R does not explain A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: The log-trig limit requires splitting the log, then expanding each factor; R describes exactly this method.

Step 1: Verify A. Split: $ln ((1 + x^{2}) cos x) = ln (1 + x^{2}) + ln (cos x)$ .

Step 2: Expand each: $ln (1 + x^{2}) \approx x^{2}$ and $ln (cos x) = ln (1 - x^{2} /2 + \dots) \approx - x^{2} /2$ .

Step 3: Sum: $x^{2} - x^{2} /2 = x^{2} /2$ . Divide by $x^{2}$ : limit $= 1/2$ . So A is true.

Step 4: Check R. The log product rule $ln (A B) = ln A + ln B$ is exactly what allows us to split and expand separately. R is true.

Step 5: Does R explain A? Yes — R provides the method (splitting) that makes the evaluation in A possible.

Trap: Students who don't split the log will struggle to expand $ln ((1 + x^{2}) cos x)$ directly and may get the wrong answer.

Generalization: Log-trig limits always require splitting before expansion.

Verification: Plugging back: numerator $\approx x^{2} /2$ , denominator $= x^{2}$ , ratio $= 1/2$ . Confirmed.

Answer: A.

Question 6 · Calculus and Optimization MCQ

Match the limits in List I with their values in List II.

\textbf{List I}

P: $lim_{x \to 0} \frac{s i n x - x}{x ^{3}}$

Q: $lim_{x \to 0} \frac{c o s x - 1}{x ^{2}}$

R: $lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}$

\textbf{List II}

1: $- 1/2$

2: $1/2$

3: $- 1/6$

A.
P-1, Q-2, R-3
B.
P-3, Q-1, R-2
C.
P-3, Q-2, R-1
D.
P-2, Q-1, R-3

Correct Answer:

Step-by-Step Solution

Key idea: Standard expansion-based limits; each requires one substitution.

Step 1 (P): $sin x = x - x^{3} /6 + \dots$

$\frac{sin x - x}{x ^{3}} = \frac{- x ^{3} /6}{x ^{3}} = - \frac{1}{6} ⟹ P-3$

Step 2 (Q): $cos x = 1 - x^{2} /2 + \dots$

$\frac{cos x - 1}{x ^{2}} = \frac{- x ^{2} /2}{x ^{2}} = - \frac{1}{2} ⟹ Q-1$

Step 3 (R): $e^{x} = 1 + x + x^{2} /2 + \dots$

$\frac{e ^{x} - 1 - x}{x ^{2}} = \frac{x ^{2} /2}{x ^{2}} = \frac{1}{2} ⟹ R-2$

Answer: P-3, Q-1, R-2 (Option B)

Question 7 · Calculus and Optimization MCQ

Match the limits in List I with their values in List II.

\textbf{List I}

P: $lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$

Q: $lim_{x \to 0} \frac{x - s i n x}{x ^{3}}$

R: $lim_{x \to 0} \frac{e ^{- x} - 1 + x - x ^{2} /2}{x ^{3}}$

\textbf{List II}

1: $1/2$

2: $1/6$

3: $- 1/6$

A.
P-1, Q-2, R-3
B.
P-1, Q-3, R-2
C.
P-2, Q-1, R-3
D.
P-3, Q-2, R-1

Correct Answer:

Step-by-Step Solution

Key idea: These are standard limits derived directly from the first few terms of the Maclaurin expansions.

Step 1 (P): $cos x = 1 - x^{2} /2 + \dots$

$\frac{1 - ( 1 - x ^{2} /2 )}{x ^{2}} = \frac{x ^{2} /2}{x ^{2}} = \frac{1}{2} ⟹ P-1$

Step 2 (Q): $sin x = x - x^{3} /6 + \dots$

$\frac{x - ( x - x ^{3} /6 )}{x ^{3}} = \frac{x ^{3} /6}{x ^{3}} = \frac{1}{6} ⟹ Q-2$

Step 3 (R): $e^{- x} = 1 - x + x^{2} /2 - x^{3} /6 + \dots$

$\frac{( 1 - x + x ^{2} /2 - x ^{3} /6 ) - 1 + x - x ^{2} /2}{x ^{3}} = \frac{- x ^{3} /6}{x ^{3}} = - \frac{1}{6} ⟹ R-3$

Answer: P-1, Q-2, R-3 (Option A)

Question 8 · Calculus and Optimization MCQ

Match the limits in List I with their values in List II.

\textbf{List I}

P: $lim_{x \to 0} \frac{l n ( 1 + x )}{x}$

Q: $lim_{x \to 0} \frac{e ^{- x} - 1}{x}$

R: $lim_{x \to 0} \frac{s i n x - x}{x ^{3}}$

\textbf{List II}

1: $1$

2: $- 1$

3: $- 1/6$

A.
P-1, Q-2, R-3
B.
P-1, Q-3, R-2
C.
P-2, Q-1, R-3
D.
P-3, Q-2, R-1

Correct Answer:

Step-by-Step Solution

Insight: These are standard limits derived directly from the first few terms of the Maclaurin expansions, testing the signs of the linear and cubic terms.

Exam route: P uses $ln (1 + x) \approx x ⟹ 1$ . Q uses $e^{- x} \approx 1 - x ⟹ - 1$ . R uses $sin x \approx x - x^{3} /6 ⟹ - 1/6$ . Match is P-1, Q-2, R-3.

Learning route:

Step 1 (P): Recall $ln (1 + x) = x - x^{2} /2 + \dots$

$\frac{ln ( 1 + x )}{x} \approx \frac{x}{x} = 1 ⟹ P-1$

Step 2 (Q): Recall $e^{u} = 1 + u + \dots$ and substitute $u = - x$ :

$e^{- x} = 1 - x + \dots$

$\frac{e ^{- x} - 1}{x} \approx \frac{- x}{x} = - 1 ⟹ Q-2$

Step 3 (R): Recall $sin x = x - x^{3} /6 + \dots$

$\frac{sin x - x}{x ^{3}} \approx \frac{- x ^{3} /6}{x ^{3}} = - \frac{1}{6} ⟹ R-3$

Answer: P-1, Q-2, R-3 (Option A)

Question 9 · Calculus and Optimization MCQ

Assertion (A): For a double series with all $a_{i, j} \geq 0$ , the two iterated sums $\sum_{i} \sum_{j} a_{i, j}$ and $\sum_{j} \sum_{i} a_{i, j}$ always give the same result (finite or infinite).

Reason (R): Non-negativity makes $∣ a_{i, j} ∣ = a_{i, j}$ , so the sum of absolute values equals the original sum, and Fubini's theorem applies whenever that common value is finite; when it is infinite both orders diverge together.

A.
Both A and R are true, and R correctly explains A.
B.
Both A and R are true, but R does not explain A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: Non-negative grids can never suffer order-dependent answers, because absolute convergence (or shared divergence) is automatic.

Step 1: Examine A. With $a_{i, j} \geq 0$ , each iterated sum computes the same total mass of the grid. Either both equal a finite number or both equal $+ \infty$ . So A is true.

Step 2: Examine R. Since $a_{i, j} \geq 0$ , $∣ a_{i, j} ∣ = a_{i, j}$ identically. Hence $\sum\sum ∣ a_{i, j} ∣ = \sum\sum a_{i, j}$ . If finite, absolute convergence holds and Fubini guarantees equality; if infinite, monotone accumulation forces both orders to $+ \infty$ . R is true.

Step 3: Does R explain A? Yes — R supplies exactly the mechanism (absolute convergence or synchronized divergence) behind A's claim.

Wrong path to watch: treating the finite/infinite distinction as breaking the argument would wrongly pick C. But the infinite case also preserves equality, so R covers everything.

Generalization: positivity removes all rearrangement risk in double sums.

Verification: take $a_{i, j} = 2^{- i} 3^{- j}$ (finite) and $a_{i, j} = \frac{1}{( i + j ) ^{2}}$ compared loosely — both orders agree in each regime.

Answer: A.

Question 10 · Calculus and Optimization MCQ

Match each limit expression in List I with its evaluation method in List II.

List I:

P: $lim_{t \to \infty} (t^{2} + t - t)$

Q: $lim_{t \to \infty} (t^{2} + 2 t - t^{2} - t)$

List II:

1: Factor out $t$ and use binomial approximation

2: Rationalize by multiplying by the conjugate

A.
P-1, Q-2
B.
P-2, Q-1
C.
P-1, Q-1
D.
P-2, Q-2

Correct Answer:

Step-by-Step Solution

Key idea: Infinity-minus-infinity limits with roots require either factoring or rationalization depending on the structure.

Step 1: Analyze P: $t^{2} + t - t$ . Factor $t$ from the root: $t^{2} + t = t 1 + 1/ t$ . Then use binomial: $1 + 1/ t \approx 1 + 1/ (2 t)$ . So $t (1 + 1/ (2 t)) - t = t + 1/2 - t = 1/2$ . Method 1 applies.

Step 2: Analyze Q: $t^{2} + 2 t - t^{2} - t$ . Both terms are roots, so rationalize: multiply by $\frac{t ^{2} + 2 t + t ^{2} - t}{t ^{2} + 2 t + t ^{2} - t}$ . Numerator becomes $(t^{2} + 2 t) - (t^{2} - t) = 3 t$ . Denominator $\approx 2 t$ for large $t$ . Limit $= 3 t /2 t = 3/2$ . Method 2 applies.

Step 3: Match: P uses method 1, Q uses method 2.

Trap: Students who try to factor $t$ from Q will get $t (1 + 2/ t - 1 - 1/ t)$ , which still requires binomial expansion and is more complex than rationalization.

Generalization: Single root minus $t$ : factor. Two roots: rationalize.

Verification: For P, factoring gives $1/2$ . For Q, rationalizing gives $3/2$ . Both methods work cleanly.

Answer: A (P-1, Q-2).