UPSC 2023 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Series of real terms: convergence, standard tests · Real Analysis · asked 2× in 13 yrs · Read the full method →

Question

Test the convergence of the series

\sum_{n=1}^\infty\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n)}\cdot\frac{x^{2n+1}}{2n+1},\quad x>0.

Ratio test for the interior; Raabe’s test for the boundary $x=1$ .

Let

a_n=\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4\cdots(2n)}\cdot\frac{x^{2n+1}}{2n+1}=\frac{(2n-1)!!}{(2n)!!}\cdot\frac{x^{2n+1}}{2n+1}.

Since $x>0$ , all $a_n>0$ .

Step 1 — Ratio test for $0<x<1$ and $x>1$ .

\frac{a_{n+1}}{a_n}=\frac{(2n+1)!!}{(2n+2)!!}\cdot\frac{(2n)!!}{(2n-1)!!}\cdot\frac{2n+1}{2n+3}\cdot x^{2}=\frac{2n+1}{2n+2}\cdot\frac{2n+1}{2n+3}\cdot x^{2}.

As $n\to\infty$ , $\frac{2n+1}{2n+2}\to 1$ and $\frac{2n+1}{2n+3}\to 1$ , so

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=x^{2}.

If $x^{2}<1$ (i.e. $0<x<1$ ): the series converges (absolutely).
If $x^{2}>1$ (i.e. $x>1$ ): the series diverges.
If $x^{2}=1$ (i.e. $x=1$ ): the ratio test is inconclusive; need a sharper analysis.

Step 2 — Boundary case $x=1$ via Raabe’s test (or direct asymptotic).

At $x=1$ ,

\frac{a_n}{a_{n+1}}=\frac{(2n+2)(2n+3)}{(2n+1)^{2}}=1+\frac{6n+5}{(2n+1)^{2}}.

Raabe’s test computes

\lim_{n\to\infty} n\!\left(\frac{a_n}{a_{n+1}}-1\right)=\lim_{n\to\infty}\frac{n(6n+5)}{(2n+1)^{2}}=\frac{6}{4}=\frac{3}{2}>1.

Since the Raabe limit $>1$ , the series converges at $x=1$ .

Step 3 — Combine.

\boxed{\;\text{The series converges for }0<x\le 1\text{ and diverges for }x>1.\;}