Question

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

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{k}}}{{\mathbf{10}}^{\mathbf{k}}}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{)}}^{\mathbf{k}}$

Short answer

The radius of convergence is$R=10$& interval is$(-5,15)$

Step-by-step solution

Ratio test

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

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

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$\sum _{k=1}^{\infty }\frac{{(-1)}^k}{{10}^k}{(x-5)}^k$

The absolute value of the ratio of the terms is$L=\left|\frac{{\displaystyle a_{k+1}}}{{\displaystyle a_k}}\right|$

Use Ratio Test$\left|\frac{{\displaystyle \frac{{\displaystyle {(-1)}^{k+1}}}{{\displaystyle {10}^{k+1}}}{(x-5)}^{k+1}}}{{\displaystyle \frac{{\displaystyle {(-1)}^k}}{{\displaystyle {10}^k}}{(x-5)}^k}}\right|=\left|\frac{{\displaystyle {(-1)}^{k+1}{(x-5)}^{k+1}}}{{\displaystyle {10}^{k+1}}}\times \frac{{\displaystyle {10}^k}}{{\displaystyle {(-1)}^k{(x-5)}^k}}\right|$

Simplify

$\begin{array}{l}=\left|\frac{(-1){(-1)}^k(x-5){(x-5)}^k}{10\cdot {10}^k}\times \frac{{10}^k}{{(-1)}^k{(x-5)}^k}\right|\\ =\left|\frac{(-1)(x-5)}{10}\times \frac{1}{1}\right|\\ =\left|\frac{(-1)}{10}(x-5)\right|\\ =\frac{|x-5|}{10}\end{array}$

For interval of convergence

From the above conclusion, if$\underset{k\to \infty }{\lim }\left|\frac{{\displaystyle a_{k+1}}}{{\displaystyle a_k}}\right|<1$then, for every$x\in R$the series$\sum _{k=1}^{\infty }\frac{{(-1)}^k}{{10}^k}{(x-5)}^k$converges.

$\begin{array}{c}\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|<1\\ \Rightarrow \frac{|x-5|}{10}<1.\end{array}$

That is,

$\begin{array}{c}|x-5|<10\\ -10<(x-5)<10\\ -5<x<15\end{array}$

So, the series converges in the interval.$(-5,15)$

Check convergence at end points

Now check the convergence of series on the point.$x=-5,15$

At.$x=-5$

$\begin{array}{c}\sum _{k=1}^{\infty }\frac{{(-1)}^k}{{10}^k}{\left(-5-5\right)}^K=\sum _{k=1}^{\infty }\frac{{(-1)}^k}{{10}^k}{\left(-10\right)}^K\\ =\sum _{k=1}^{\infty }\frac{{(-1)}^k{(-1)}^k}{{10}^k}\\ =\sum _{k=1}^{\infty }{(-1)}^{2k}\\ =\sum _{k=1}^{\infty }1\end{array}$

So, diverges at$x=-5$

At.$x=15$

$\begin{array}{c}\sum _{k=1}^{\infty }\frac{{(-1)}^k}{{10}^k}{\left(15-5\right)}^K=\sum _{k=1}^{\infty }\frac{{(-1)}^k}{{10}^k}{\left(10\right)}^K\\ =\sum _{k=1}^{\infty }{(-1)}^k\end{array}$

So, $\sum _{k=1}^{\infty }{(-1)}^k$diverges. Hence, the interval of convergence is.$(-5,15)$

Radius of convergence

The radius of the convergence is,

$\begin{array}{c}R=\frac{15-(-5)}{2}\\ =\frac{15+5}{2}\\ =\frac{20}{2}\\ =10\end{array}$

Hence, the required radius of the convergence is.$R=10$