← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Absolute and conditional convergence · Real Analysis · asked 4× in 13 yrs · Read the full method →

Question

Find the range of $p(>0)$ for which the series

is (i) absolutely convergent and (ii) conditionally convergent.

Technique

Limit comparison / integral test for absolute convergence ($p$-series threshold $p>1$); Leibniz test for plain convergence (holds for all $p>0$); subtract to get the conditional range.

Solution

Write the series as

Step 1 — Absolute convergence

The series of absolute values is

Since $a>0$ is fixed, $n+a\sim n$ as $n\to\infty$, so by the limit comparison test with the $p$-series $\sum 1/n^p$:

Hence $\sum \dfrac{1}{(n+a)^p}$ converges iff $\sum \dfrac1{n^p}$ converges, i.e. iff $p>1$. (Equivalently, by the integral test, $\int_1^\infty (x+a)^{-p}dx$ converges iff $p>1$.) Therefore

Step 2 — Convergence of the alternating series (for the conditional case)

Consider the terms $b_n=\dfrac{1}{(n+a)^p}>0$. For every $p>0$ and $a>0$:

• $b_n>0$;
• $b_n$ is decreasing: $n+a<(n+1)+a\Rightarrow (n+a)^p<(n+1+a)^p\Rightarrow b_{n+1}<b_n$;
• $b_n\to0$ as $n\to\infty$ (since $(n+a)^p\to\infty$ for $p>0$).

By the Leibniz (alternating series) test, $\sum(-1)^{n+1}b_n$ converges for every $p>0$.

Step 3 — Conditional convergence

A series is conditionally convergent iff it converges but does not converge absolutely. By Step 2 the series converges for all $p>0$; by Step 1 it converges absolutely iff $p>1$. Hence it converges conditionally exactly when it converges but not absolutely:

(For $p>1$ it is absolutely convergent — not “conditionally”; for $0<p\le1$ it converges by Leibniz but $\sum 1/(n+a)^p$ diverges, so the convergence is conditional. There is no $p>0$ for which the series diverges.)

Summary

Answer