GATE 2026 EC Q.36: solution and answer

GATE 2026 · ECQ.36 · 2 marks · MCQ

Consider the two series, $S_{A}$ and $S_{B}$ , where

S_{A} = n = 1 \sum \infty \frac{n ^{2}}{2 ^{n}}

S_{B} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \dots

Which of the following statements is correct for the two given series?

(A)
Both $S_{A}$ and $S_{B}$ converge.
(B)
Neither $S_{A}$ nor $S_{B}$ converges.
(C)
$S_{A}$ converges but $S_{B}$ does not converge.
(D)
$S_{B}$ converges but $S_{A}$ does not converge.

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Answer and solution

ANSWER

(A) Both $S_{A}$ and $S_{B}$ converge ( $S_{A} = 6$ , $S_{B} = \frac{12}{7}$ ).

Official answer: ASource: Official answer key, GATE 2026 EC (IIT Guwahati)

$S_{A}$ , ratio test. With $a_{n} = \frac{n ^{2}}{2 ^{n}}$ : $\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 ) ^{2}}{2 ^{n + 1}} \cdot \frac{2 ^{n}}{n ^{2}} = \frac{1}{2} (1 + \frac{1}{n})^{2} \to \frac{1}{2} < 1,$ so $S_{A}$ converges.
Check with the standard sum $\sum_{n = 1}^{\infty} n^{2} x^{n} = \frac{x ( 1 + x )}{( 1 - x ) ^{3}}$ for $∣ x ∣ < 1$ . At $x = \frac{1}{2}$ : $S_{A} = \frac{\frac{1}{2} \cdot \frac{3}{2}}{( \frac{1}{2} ) ^{3}} = \frac{3}{4} \cdot 8 = 6,$ which is finite.
$S_{B}$ , spot the pattern. The terms are $2^{0}, 2^{- 1}, 2^{- 3}, 2^{- 4}, 2^{- 6}, 2^{- 7}, 2^{- 9}, \dots$ : powers of $2$ in pairs, with every third power ( $2^{- 2}, 2^{- 5}, 2^{- 8}, \dots$ ) left out.
Comparison test. Every term of $S_{B}$ is positive, and the terms are distinct terms of the geometric series $\sum_{k = 0}^{\infty} 2^{- k} = 2$ . So the partial sums of $S_{B}$ increase and never exceed $2$ . A bounded increasing sequence converges, so $S_{B}$ converges.
Exact value (a check, not needed for the answer): since all terms are positive, grouping consecutive pairs does not change the sum. The $k$ -th pair ( $k = 0, 1, 2, \dots$ ) is $2^{- 3 k} (1 + \frac{1}{2}) = \frac{3}{2} \cdot 8^{- k}$ , so $S_{B} = \frac{3}{2} (1 + \frac{1}{8} + \frac{1}{64} + \dots) = \frac{3}{2} \cdot \frac{1}{1 - \frac{1}{8}} = \frac{3}{2} \cdot \frac{8}{7} = \frac{12}{7} .$
Both series converge, so option (A) is correct.

KEY IDEA

An exponential in the denominator beats any power of $n$ in the numerator: for $\sum n^{k} / r^{n}$ with $r > 1$ the ratio test gives the limit $1/ r < 1$ , so the series converges. A series of positive terms, each a distinct term of a convergent geometric series, has bounded increasing partial sums and therefore converges, however irregular the pattern of left-out terms.

WATCH OUT FOR

Two traps. First, seeing $n^{2}$ in the numerator of $S_{A}$ and deciding the terms grow, when $2^{n}$ grows much faster and the terms tend to $0$ . Second, deciding that $S_{B}$ is not geometric because of the gaps ( $\frac{1}{4}, \frac{1}{32}, \dots$ are missing) and so it cannot be judged. Leaving out positive terms from a convergent geometric series can only make the sum smaller, so it still converges.

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GATE 2026 EC Q.36: solution and answer

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