Q 51. - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

Step 2. Find the radius of convergence.

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

Ratio for the absolute convergence is 

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

Since, for $k\to \infty$The limit is zero irrespective  of the value of variable.

$\underset{k\to \infty }{\lim }\left(\frac{k+1}{\left(k+m\right)\left(k+m\right)}\right)\left|x\right|=0$

So, by the ratio test the series converges absolutely for every value of $x$.

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