Question

Find the radius of convergence for the given series:$\sum _{k=0}^{\infty }\frac{k!}{{\left(\left(k+m\right)!\right)}^2}{x}^k$

Short answer

The radius of convergence for the series is$1$.

Step-by-step solution

Step 1. Given information.  

The given power series is$\sum _{k=0}^{\infty }\frac{k!}{{\left(\left(k+m\right)!\right)}^2}$.

Step 2. Find the radius of convergence. 

Let us take$b_k=\frac{k!}{{\left(\left(k+m\right)!\right)}^2}{x}^k$therefore$b_{k+1}=\frac{\left(k+1\right)!}{{\left(\left(k+1+m\right)!\right)}^2}{x}^{k+1}$

Thus, $$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle \frac{\left(k+1\right)!}{{\left(\left(k+1+m\right)!\right)}^2}}{x}^{k+1}}{{\displaystyle \frac{k!}{{\left(\left(k+m\right)!\right)}^2}}{x}^k}\right| \\ =\underset{k\to \infty }{\lim }\left|\left(\frac{k+1}{{\left(k+1+m\right)}^2}\right)x\right| \end{gathered}$$

So, by the ratio test of absolute convergence, the series will converge when $\left|x\right|<1$.

This implies that $x\in (-1,1)$.

Therefore, the radius of the convergence for the series is$1$.