How to find the interval of convergence for this series

Question

How to find the interval of convergence for this series.

$\sum _{ n=0 }^{ \infty }{ \frac { 2n(x+1)^ n }{ 3^ n } }$

My attempt

I used ratio test and got the radius of convergence as $\frac { -4 }{ 3 } <x<\frac { -2 }{ 3 } $

After this I substituted x=$\frac { -4 }{ 3 }$ and x=$\frac { - 2}{ 3 }$

I don't understand what to do after substituting the values.. can anyone show how to find the interval of convergence

Answer 1

Let's first double check. $$\left|\frac{\frac{2(n+1)(x+1)^{n+1}}{3^{n+1}}}{\frac{2n(x+1)^n}{3^n}}\right|=\left|\frac{(n+1)(x+1)}{3n}\right|\;.$$ So $\lim_{n\to \infty}\left|\frac{(n+1)(x+1)}{3n}\right|=\frac{1}{3}\vert x+1\vert<1$ if and only if $$-3<x+1<3\Leftrightarrow -4<x<2\;.$$ But if we want to know if the series converges for $x=-4$ and $x=2$, we've got to check directly.

For $x=-4$, we have $$\sum_{n=0}^{\infty}\frac{-2*3^n*n}{3^n}=\sum_{n=0}^{\infty}-2n<-\infty\;,$$ and for $x=2$, we have $$\sum_{n=0}^{\infty}\frac{2*3^n*n}{3^n}=\sum_{n=0}^{\infty}2n>\infty$$

So the interval of convergence is $(-4,2)$.

Answer 2

You made an error somewhere. Note that\begin{align}\lim_{n\to\infty}\left|\frac{\frac{2(n+1)}{3^{n+1}}(x+1)^{n+1}}{\frac{2n}{3^n}(x+1)^n}\right|&=\lim_{n\to\infty}\frac{n+1}{3n}|x+1|\\&=\frac{|x+1|}3.\end{align}Therefore, the series converges when $|x+1|<3$ and diverges when $|x+1|>3$. So, the radius of convergence is $3$.

If $|x+1|=3$, then $\left|\frac{2n}{3^n}(x+1)^n\right|=2n$. Since you don't have $\lim_{n\to\infty}2n=0$, your series diverges when $|x+1|=3$. Therefore, the interval of convergence is $(-4,2)$.