Question

Find the radius of convergence and interval of convergence of the series. (enter your answer as an improper fraction.) sum_(n = 1)^infinity (2^n (x + 8)^n)/sqrt(n)

Answer 1

$\displaystyle\sum_{n=1}^\infty\frac{2^n(x+8)^n}{\sqrt n}$

Ratio test:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}(x+8)^{n+1}}{\sqrt{n+1}}}{\frac{2^n(x+8)^n}{\sqrt n}}\right|=|x+8|\lim_{n\to\infty}\frac{2\sqrt n}{\sqrt{n+1}}=2|x+8|\lim_{n\to\infty}\sqrt{\frac n{n+1}}$

It's easy to show the remaining limit is 1.

The ratio test says that if this limit is less than 1, then the series converges. So the series will converge for any $x$ satisfying

$2|x+8|<1\implies|x+8|<\dfrac12$

which means the radius of convergence is $\dfrac12$, and the interval of convergence is $-\dfrac{17}2<x<-\dfrac{15}2$.

We're not finished, though, as we still need to check the endpoints of the interval.

When $x=-\dfrac{17}2$, we have

$\displaystyle\sum_{n=1}^\infty\frac{2^n\left(-\frac12\right)^n}{\sqrt n}=\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n}$

which converges by the alternating series test, as

$\bullet$ $\displaystyle\lim_{n\to\infty}\frac1{\sqrt n}=0$

$\bullet$ $\dfrac1{\sqrt n}$ is strictly decreasing for $n\in\mathbb N$

On the other hand, when $x=-\dfrac{15}2$, we get

$\displaystyle\sum_{n=1}^\infty\frac{2^n\left(\frac12\right)^n}{\sqrt n}=\sum_{n=1}^\infty\frac1{\sqrt n}$

which is a divergent $p$-series.

Therefore the complete interval of convergence is $-\dfrac{17}2\le x<-\dfrac{15}2$.