Question

In Problems 1-10 find the interval and radius of convergence for the given power series.

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathit{k}\mathbf{!}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}$

Short answer

The radius of convergence is$R=0$& converges for only.$x=1$

Step-by-step solution

Ratio test

The Ratio test for the power series, $\sum _{n=1}^{\infty }{a}_n{(x-a)}^n$is given as,

$\underset{n\to \infty }{lim}\left|\frac{{\displaystyle a_{n+1}{(x-a)}^{n+1}}}{{\displaystyle a_n{(x-a)}^n}}\right|=L$

If,$L<1$ then the series converges absolutely.

If,$L>1$then the series diverges.

If,L= 1 then the test is inconclusive.

Apply ratio test

Consider the series$\sum _{k=0}^{\infty }k!{(x-1)}^k$

Let$a_k=k!{(x-1)}^k$and$a_{k+1}=(k+1)!{(x-1)}^{k+1}$

Then by using the ratio test,

$\begin{array}{c}L=\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|\text{Useratiotest}\\ =\underset{k\to \infty }{\lim }\left|\frac{(k+1)!{(x-1)}^{k+1}}{k!{(x-1)}^k}\right|\text{Substitutetheterms}\\ =\underset{k\to \infty }{\lim }\left(\right|(k+1)\cdot (x-1)\left|\right)\text{Simplify}\\ L=\left|\right(x-1\left)\right|\underset{k\to \infty }{\lim }\left(\right|k+1\left|\right)\text{Take}(x-1)\text{asoutsideofthelimit}=\\ =\infty \end{array}$

Find radius & interval of convergence

From the above $L>1$for every,$x\in R$ the series$\sum _{k=0}^{\infty }k!{(x-1)}^k$diverges.

Check the convergence at points where.$k!{(x-1)}^k=0$

That means$x=1$

So the series converges only for$x=1$

Thus the radius of convergence=0

Hence, the series converges only for.$x=1$