Question Bank • QB_EM_ADV_038 NAT • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Sequences and series – infinite series, tests for convergence

Question Statement:

The radius of convergence $R$ of the power series $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} x^n$ is ________ (round off to two decimal places).

Answer Key:

0.25

Step-by-Step Solution & Explanation

**Method 1: D'Alembert's Ratio Test** Let $a_n = \frac{(2n)!}{(n!)^2}$. The ratio of consecutive coefficients is: $\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!} = \frac{(2n+2)(2n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1}$ Taking the limit as $n \to \infty$: $L = \lim_{n \to \infty} \frac{4n + 2}{n + 1} = 4$ The radius of convergence $R$ is the reciprocal of $L$: $R = \frac{1}{L} = \frac{1}{4} = 0.25$ **Method 2: Stirling's Approximation Shortcut** Using $n! \sim \sqrt{2\pi n} (n/e)^n$: $a_n = \binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}} \implies \lim_{n \to \infty} (a_n)^{1/n} = 4 \implies R = \frac{1}{4} = 0.25$
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