Find Simple Convergent sum We have a sum
$\sum_{0}^{\infty} 2^{n+1}(x+1)^{3n+1}$
I am asked to find all values of x s.t this sum converges then compute the sum.
I have no idea what im doing so i did this. $\frac {2^{n}(x+1)^{3n}} {2^{n+1}(x+1)^{3n+1}}$
this yields $\frac {1}{2(x+1)^{3}}$ this looks alittle wierd but i just move it over and solved to get $x=-1 +2^{-1/3}$ this isnt right i also have no idea to find the sum if it was please help.

First you should realize that just about the only series you've learned how to get an explicit sum for is a geometric series. So your first goal should be to make the given series looks like the standard form of a geometric series. One way to do this is as follows:
\begin{align}
\sum_{n = 0}^\infty 2^{n+1} (x+1)^{3n+1} & = 2(x+1) \sum_{n = 0}^\infty 2^{n} (x+1)^{3n} \\
 & = 2(x+1) \sum_{n = 0}^\infty (2(x+1)^3)^n.
\end{align}
Now $\displaystyle \sum_{n = 0}^\infty (2(x+1)^3)^n$ is a geometric series with $a = 1$ and $r = 2(x+1)^3$. Hence
$$\sum_{n = 0}^\infty (2(x+1)^3)^n = \frac{1}{1 - 2(x+1)^3},$$
and from the work about it follows that
$$\sum_{n = 0}^\infty 2^{n+1} (x+1)^{3n+1} = 2(x+1) \sum_{n = 0}^\infty (2(x+1)^3)^n = \frac{2(x + 1)}{1 - 2(x+1)^3}.$$
A geometric series $\displaystyle \sum_{n=0}^\infty ar^n$ converges if and only if $|r| < 1$, so our given series converges if and only if
$$|2(x+1)^3| < 1 \iff -1 - 2^{-1/3} < x < -1 + 2^{-1/3}.$$
