UPSC 2025 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Absolute and conditional convergence · Real Analysis · asked 4× in 14 yrs · Read the full method →

Question

Examine whether the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n}$ is absolutely or conditionally convergent.

Technique

Use the Leibniz alternating series test to establish convergence, and the divergence of the harmonic series ( $p$ -series with $p=1$ ) to rule out absolute convergence.

Solution

Write $\displaystyle\sum_{n=1}^{\infty} a_n$ with $a_n = \dfrac{(-1)^{n-1}}{n} = 1 - \tfrac12 + \tfrac13 - \tfrac14 + \cdots$ .

Step 1 — Test for absolute convergence.

Absolute convergence means $\displaystyle\sum_{n=1}^{\infty} |a_n|$ converges. Here

\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \frac{1}{n},

the harmonic series. This is a $p$ -series with $p=1$ , which diverges (by the integral test, $\sum_{n=1}^{N}\frac1n \ge \int_1^{N+1}\frac{dx}{x} = \ln(N+1)\to\infty$ ).

Therefore the series is not absolutely convergent.

Step 2 — Test for convergence by Leibniz’s test.

The series is alternating: $a_n = (-1)^{n-1} b_n$ with $b_n = \dfrac1n > 0$ . Leibniz’s test requires:

Monotone decreasing: $b_{n+1} = \dfrac{1}{n+1} < \dfrac1n = b_n$ for all $n\ge1$ .
Limit zero: $\displaystyle\lim_{n\to\infty} b_n = \lim_{n\to\infty}\frac1n = 0$ .

Both conditions hold, so by Leibniz’s test the series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ converges.

Step 3 — Conclusion.

The series converges but does not converge absolutely. By definition, it is conditionally convergent. (Its sum is in fact $\ln 2$ .)

Answer

The series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ is conditionally convergent: it converges by Leibniz’s test, but $\sum 1/n$ diverges, so it is not absolutely convergent. Its sum is $\ln 2 \approx 0.6931$ .