GATE 2025 EC Q.12: solution and answer

GATE 2025 · ECQ.12 · 1 mark · MCQ

Consider the following series:

(i) $\sum_{n = 1}^{\infty} \frac{1}{n}$

(ii) $\sum_{n = 1}^{\infty} \frac{1}{n ( n + 1 )}$

(iii) $\sum_{n = 1}^{\infty} \frac{1}{n !}$

Choose the correct option.

(A)
Only (ii) converges
(B)
Only (ii) and (iii) converge
(C)
Only (iii) converges
(D)
All three converge

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Answer and solution

ANSWER

(B) Only (ii) and (iii) converge. Series (i) is a p-series with p = 1/2, so it diverges. Series (ii) telescopes to 1. Series (iii) passes the ratio test and sums to e − 1.

Official answer: BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Series (i), p-series test: $\sum \frac{1}{n ^{p}}$ converges only when $p > 1$ . Here $\frac{1}{n} = \frac{1}{n ^{1/2}}$ , so $p = \frac{1}{2} \leq 1$ and (i) diverges. Its terms do go to zero, but that alone does not make a series converge.
Series (ii), partial fractions: $\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$ .
Telescoping partial sum: $S_{N} = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{N} - \frac{1}{N + 1}) = 1 - \frac{1}{N + 1} \to 1$ so (ii) converges (to $1$ ). The limit comparison test with $\sum \frac{1}{n ^{2}}$ ( $p = 2 > 1$ ) gives the same result, since $\frac{1/ ( n ( n + 1 ))}{1/ n ^{2}} = \frac{n}{n + 1} \to 1$ .
Series (iii), ratio test: $\frac{a _{n + 1}}{a _{n}} = \frac{n !}{( n + 1 )!} = \frac{1}{n + 1} \to 0 < 1$ , so (iii) converges. From the series for $e^{x}$ at $x = 1$ (with the $n = 0$ term, $1$ , left out), its sum is $e - 1$ .
Only (ii) and (iii) converge, which is option (B).

KEY IDEA

Use a standard test for each series. For $\sum 1/ n^{p}$ , check whether $p > 1$ . For $\sum 1/ (n (n + 1))$ , split into partial fractions so the sum telescopes. For $\sum 1/ n!$ , the ratio test works well because consecutive factorials cancel down to $1/ (n + 1)$ .

WATCH OUT FOR

Deciding that (i) converges because $1/ n \to 0$ . A term going to zero is necessary for convergence but not sufficient. $\sum 1/ n$ is a p-series with $p = \frac{1}{2} \leq 1$ , so it diverges, and choosing option (D) this way is wrong. A second slip is to confuse $p = \frac{1}{2}$ with the convergent case and treat every series whose terms shrink as convergent.

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GATE 2025 EC Q.12: solution and answer

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