Sequences, Series and Limits PYQs for GATE DA

Solve 3+ Sequences, Series and Limits previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Sequences, Series and Limits

Chapter Roadmap

Sequences, Series and Limits

1. Infinite Series and Double Summations

Current Topic | Focus: Convergence criteria, exact sums, iterated 2D grids

2. Limits Using Algebraic and Logarithmic Expansions

Next Topic | Focus: Taylor series, L'Hopital's rule, asymptotic behavior

Topic Hero: The Intuition of Infinite Addition

The Intuition of Infinite Addition

The Core Question:

Does adding infinitely many terms yield a finite number?

1D Infinite Series

Summing along a single sequence: $S = \sum_{n = 1}^{\infty} a_{n}$

Convergent: Partial sums approach a finite limit $L$ .
Divergent: Partial sums grow without bound or oscillate.

2D Double Summations

Summing over a grid of indices $(i, j)$ : $S = \sum_{i} \sum_{j} a_{i, j}$

Iterated Sum: Sum one row completely, then add the row sums.
The Big Question: Does row-by-row equal column-by-column?

Sequences, Series and Limits: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Calculus and Optimization · 2026 NAT

The value of $\sum_{i = 0}^{\infty} \sum_{j = 1}^{\infty} 2^{- i} 3^{- j}$ is ______________ . (Answer in integer)

Correct Answer:

1.00

Step-by-Step Solution

Insight: The summand $2^{- i} 3^{- j}$ is a product of a function of $i$ alone and a function of $j$ alone, so the double sum separates into the product of two independent geometric series.

Exam route: Split $\sum_{i} \sum_{j} 2^{- i} 3^{- j} = (\sum_{i = 0}^{\infty} 2^{- i}) (\sum_{j = 1}^{\infty} 3^{- j})$ . Evaluate each geometric series using $S = \frac{a}{1 - r}$ , then multiply.

Learning route:

This is a separable double summation question, recognisable because the general term $a_{i, j} = 2^{- i} \cdot 3^{- j}$ factors cleanly into $f (i) \cdot g (j)$ .

Step 1 — Apply the separation rule (Fubini for positive terms):

$i = 0 \sum \infty j = 1 \sum \infty 2^{- i} 3^{- j} = (i = 0 \sum \infty (\frac{1}{2})^{i}) \times (j = 1 \sum \infty (\frac{1}{3})^{j})$

Step 2 — First geometric series ( $i$ starts at $0$ , ratio $r = \frac{1}{2}$ ):

$S_{1} = \frac{1}{1 - \frac{1}{2}} = 2$

Step 3 — Second geometric series ( $j$ starts at $1$ , ratio $r = \frac{1}{3}$ , first term $a = \frac{1}{3}$ ):

$S_{2} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

Step 4 — Multiply:

$S = 2 \times \frac{1}{2} = 1$

Trap check: A common mistake is to start the $j$ -sum at $0$ instead of $1$ , which would give $S_{2} = \frac{1}{1 - 1/3} = \frac{3}{2}$ and a wrong total of $3$ . Always read the lower limit carefully.

Verification: Evaluate the inner sum first: $\sum_{j = 1}^{\infty} 2^{- i} 3^{- j} = 2^{- i} \cdot \frac{1}{2}$ . Then $\sum_{i = 0}^{\infty} 2^{- i} \cdot \frac{1}{2} = \frac{1}{2} \cdot 2 = 1$ . Confirmed.

The answer is $1$ .

Question 2 · Calculus and Optimization · 2025 NAT

$t \to + \infty lim t^{2} + t - t =$

(Round off to one decimal place)

Correct Answer:

0.50

Step-by-Step Solution

Insight: This is an $\infty - \infty$ indeterminate form. Rationalise by multiplying and dividing by the conjugate $t^{2} + t + t$ , or equivalently factor out $t$ and use the binomial approximation.

Exam route: Rationalise → simplify → divide numerator and denominator by $t$ → substitute $t \to \infty$ .

Learning route:

This is an infinity-minus-infinity limit question, recognisable because both $t^{2} + t \to \infty$ and $t \to \infty$ , giving the indeterminate form $\infty - \infty$ .

Method 1 — Conjugate rationalisation (recommended for exams):

Step 1 — Multiply and divide by the conjugate:

$t^{2} + t - t = \frac{( t ^{2} + t - t ) ( t ^{2} + t + t )}{t ^{2} + t + t}$

Step 2 — Simplify the numerator using $(a - b) (a + b) = a^{2} - b^{2}$ :

$= \frac{( t ^{2} + t ) - t ^{2}}{t ^{2} + t + t} = \frac{t}{t ^{2} + t + t}$

Step 3 — Divide numerator and denominator by $t$ (valid since $t > 0$ ):

$= \frac{1}{1 + \frac{1}{t} + 1}$

Step 4 — Take the limit as $t \to \infty$ (so $1/ t \to 0$ ):

$L = \frac{1}{1 + 0 + 1} = \frac{1}{2}$

Method 2 — Binomial expansion:

Factor $t$ from the square root: $t^{2} + t = t 1 + 1/ t$ .

Using $(1 + u)^{1/2} \approx 1 + u /2$ with $u = 1/ t$ :

$t (1 + \frac{1}{2 t}) - t = t + \frac{1}{2} - t = \frac{1}{2}$

Trap check: A common error is to conclude the limit is $0$ by arguing "both terms go to infinity, so they cancel." This ignores the sub-leading $1/2$ term that survives.

Verification: At $t = 1000$ : $1000000 + 1000 - 1000 = 1001000 - 1000 \approx 1000.4999 - 1000 = 0.4999 \approx 0.5$ . Confirmed.

The answer is $0.5$ (rounded to one decimal place: $0.5$ ).

Question 3 · Calculus and Optimization · 2024 NAT

Evaluate the following limit:

x \to 0 lim \frac{ln (( x ^{2} + 1 ) cos x )}{x ^{2}} =______.

Correct Answer:

0.50

Step-by-Step Solution

Insight: The numerator is a log of a product, so split it into $ln (1 + x^{2}) + ln (cos x)$ , then expand each piece to order $x^{2}$ using standard Taylor series.

Exam route: Split log → expand $ln (1 + x^{2}) \approx x^{2}$ and $ln (cos x) \approx - x^{2} /2$ → add → divide by $x^{2}$ → get $\frac{1}{2}$ .

Learning route:

This is a log-trig limit via Taylor expansion question, recognisable because direct substitution gives $\frac{l n ( 1 \cdot 1 )}{0} = \frac{0}{0}$ , an indeterminate form, and the numerator contains $ln$ of a product involving a trig function.

Step 1 — Use the log product rule:

$ln ((x^{2} + 1) cos x) = ln (1 + x^{2}) + ln (cos x)$

Step 2 — Expand $ln (1 + x^{2})$ using $ln (1 + u) = u - \frac{u ^{2}}{2} + \dots$ with $u = x^{2}$ :

$ln (1 + x^{2}) = x^{2} - \frac{x ^{4}}{2} + \dots \approx x^{2} (to order x^{2})$

Step 3 — Expand $ln (cos x)$ . First, $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ . Let $u = - \frac{x ^{2}}{2} + \dots$ , then:

$ln (cos x) = ln (1 + u) \approx u = - \frac{x ^{2}}{2} (to order x^{2})$

Step 4 — Combine the numerator:

$ln (1 + x^{2}) + ln (cos x) \approx x^{2} - \frac{x ^{2}}{2} = \frac{x ^{2}}{2}$

Step 5 — Divide by $x^{2}$ :

$L = x \to 0 lim \frac{x ^{2} /2}{x ^{2}} = \frac{1}{2}$

Trap check: If you forget to expand $ln (cos x)$ and only keep $ln (1 + x^{2}) \approx x^{2}$ , you would get $L = 1$ , which is wrong. Both log terms contribute at order $x^{2}$ .

Verification: Numerical check at $x = 0.01$ : $ln ((0.0001 + 1) cos 0.01) /0.0001 \approx ln (1.0001 \times 0.99995) /0.0001 \approx ln (1.00005) /0.0001 \approx 0.5000$ . Confirmed.

The answer is $0.50$ .

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Sequences, Series and Limits PYQs for GATE DA

Chapter Roadmap: Sequences, Series and Limits

Chapter Roadmap

1. Infinite Series and Double Summations

2. Limits Using Algebraic and Logarithmic Expansions

Topic Hero: The Intuition of Infinite Addition

The Intuition of Infinite Addition

1D Infinite Series

2D Double Summations

Sequences, Series and Limits: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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