Question

Prove that the power series \(\sum_{n=0}^{\infty} \frac{(n+p) !}{n !(n+q) !} x^{n}\) has a radius of convergence of \(R=\infty\) if \(p\) and \(q\) are positive integers.

Short answer

\(R = \infty\). When \(|x| = 0\), the series \(\sum_{n=0}^{\infty} \frac{(n+p) !}{n !(n+q) !} x^{n}\) converges absolutely which means the radius of convergence is \(\infty\).

Step-by-step solution

Write down the terms of the series

First, write down the formulas for \(a_n\) and \(a_{n+1}\). We have \(a_n = \frac{(n+p)!}{n!(n+q)!}x^n\) and \(a_{n+1} = \frac{(n+1+p)!}{(n+1)!(n+1+q)!}x^{n+1}\) which represent the nth and (n+1)st terms of the series respectively.

Compute Ratio and Apply Ratio Test

Next, compute the ratio \(\frac{a_{n+1}}{a_n} = \frac{(n+1+p)!}{(n+1)!(n+1+q)!}x^{n+1} \cdot \frac{n!(n+q)!}{(n+p)!x^n} = \frac{(n+1+p)(n+q)!x}{(n+1+q)(n+p)!}\). Now calculate the limit of this ratio as n approaches infinity, \(\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n\rightarrow\infty} |\frac{(n+1+p)(n+q)!x}{(n+1+q)(n+p)!}|\). As n approaches infinity, the term \((n+1+p)/(n+1+q)\) approaches 1. This implies that the limit as n approaches infinity is simply |x|. Since the radius of convergence is defined as the reciprocal of this limit, and since p and q are positive integers, the radius of convergence \(R = \frac{1}{|x|} = \infty\) when \(|x| = 0\).

Statement of Result

Since the value of \(|x|\) (the modulo of x) for which the series \(\sum_{n=0}^{\infty} \frac{(n+p) !}{n !(n+q) !}x^{n}\) converges absolutely is zero, and all other x within the range of the series, hence the radius of convergence \(R = \infty\).