Question

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

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{5}\mathbf{k}}}{{\mathbf{5}}^{\mathbf{2}\mathbf{k}}}{\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{3}}\mathbf{\right)}}^{\mathbf{k}}$

Short answer

The radius of convergence is.$R=\frac{75}{32}\text{interval:}(-\frac{75}{32},\frac{75}{32})$

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$\sum _{k=1}^{\infty }\frac{2^{5k}}{5^{2k}}{\left(\frac{x}{3}\right)}^k$

By ratio test we have:

$\begin{array}{c}\underset{k\to \infty }{\lim }\left|\frac{\frac{2^{5(k+1)}{x}^{(k+1)}}{5^{2(k+1)}{3}^{(k+1)}}}{\frac{2^{5k}{x}^k}{5^{2k}{3}^k}}\right||=\underset{k\to \infty }{\lim }\left|\frac{2^{5(k+1)}{x}^{(k+1)}}{5^{2(k+1)}{3}^{(k+1)}}\cdot \frac{5^{2k}{3}^k}{2^{5k}{x}^k}\right|\\ =\underset{k\to \infty }{\lim }\left|\frac{2^{kk}\cdot 2^5\cdot 3k\cdot 5^{2k}\cdot x\cdot x}{2^{kk}\cdot 3\cdot 3\cdot 5^{2k}\cdot 5^2\cdot x}\right|\\ =\underset{k\to \infty }{\lim }\left|\frac{2^{5}x}{3\cdot 5^2}\right|\\ =\frac{2^5}{3\cdot 5^2}\left|x\right|\end{array}$

Find radius & interval of convergence

Now we have to determine the interval and radius of convergence.

$\begin{array}{c}\text{If}\frac{2^5}{3\cdot 5^2}\left|x\right|<1\\ \Rightarrow \left|x\right|<\frac{75}{32}\end{array}$

We observe that the power series is centered at $a=0$and the radius.$R=\frac{75}{32}$

And the interval of convergence is$(0-\frac{75}{32},0+\frac{75}{32})$

Thus the interval is.$(-\frac{75}{32},\frac{75}{32})$

Check convergence at end points

The endpoints are$x=-\frac{75}{32}$and.$x=\frac{75}{32}$

For the left end point,$x=-\frac{75}{32}$ the series becomes:

$\begin{array}{c}\sum _{k=1}^{\infty }\frac{2^{5k}}{5^{2k}}{(-\frac{75}{32}\cdot \frac{1}{3})}^k=\sum _{k=1}^{\infty }\frac{2^{5k}}{5^{2k}}{(-\frac{25}{32})}^k\\ =\sum _{k=1}^{\infty }{(-\frac{2^5\cdot 25}{5^2\cdot 32})}^k\\ =\sum _{k=1}^{\infty }{(-1)}^k\text{diverges}\end{array}$

At, $x=\frac{75}{32}$the series becomes

$\begin{array}{c}\sum _{k=1}^{\infty }\frac{2^{5k}}{5^{2k}}{(\frac{75}{32}\cdot \frac{1}{3})}^k=\sum _{k=1}^{\infty }\frac{2^{5k}}{5^{2k}}{\left(\frac{25}{32}\right)}^k\\ =\sum _{k=1}^{\infty }{\left(\frac{2^5\cdot 25}{5^2\cdot 32}\right)}^k\\ =\sum _{k=1}^{\infty }{\left(1\right)}^k\text{diverges\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(sincetheconstantseriesisdiverges)}\end{array}$

Thus the radius and interval of convergence is.$R=\frac{75}{32}\text{interval:}(-\frac{75}{32},\frac{75}{32})$