Q 52. - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

Step 2. Find the radius of convergence.

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

Ratio for the absolute convergence is 

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

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

Thus,

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

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

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