Question

Find the interval of convergence of the power series

$\sum _{k=1}^{\infty }\frac{k!}{k^k}(x-4{)}^k$

Short answer

The interval of convergence of the power series $\sum _{k=1}^{\infty }\frac{k!}{k^k}(x-4{)}^k$is $(3,5)$

Step-by-step solution

Given information

The power series is$\sum _{k=1}^{\infty }\frac{k!}{k^k}(x-4{)}^k$

The ratio test for absolute convergence will be used to determine the convergence interval.

Let, the first assume $b_k=\frac{k!}{k^k}(x-4{)}^k$, therefore $b_{k+1}=\frac{(k+1)!}{(k+1{)}^{k+1}}(x-4{)}^{k+1}$

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{\frac{(k+1)!}{(k+1{)}^{k+1}}(x-4{)}^{k+1}}{\frac{k!}{k^k}(x-4{)}^k}\right| \\ =\underset{k\to \infty }{\lim }\frac{(k+1)k^k}{(k+1{)}^{k+1}}|x-4| \\ =\underset{k\to \infty }{\lim }{\left(\frac{k}{k+1}\right)}^k|x-4| \end{gathered}$$

The limit is $|x-4|$

The ratio test for absolute convergence will be used to determine the convergence interval when $|x-4|<1$that is $-1<x-4<1$

As a result, we may write $-1<x-4$and $x-4<1$

Implies that

$x>3andx<5$

$x\in (3,5)$

Now, because the intervals are limited, we examine the series' behavior at the ends.

When $x=3$

$$\begin{gathered} {\left.\sum _{k=1}^{\infty }\frac{k!}{k^k}(x-4{)}^k\right|}_{x=5}=\sum _{k=1}^{\infty }\frac{k!}{k^k}(5-4{)}^k \\ =\sum _{k=1}^{\infty }\frac{k!}{k^k}(1{)}^k \\ =\sum _{k=1}^{\infty }\frac{k!}{k^k} \end{gathered}$$

The constant multiple, which diverges, is the consequence.

The interval of convergence of the power series

The interval of convergence of the power series $\sum _{k=1}^{\infty }\frac{k!}{k^k}(x-4{)}^k$is$(3,5)$