<s>[INST] 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.
 [/INST] 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
$$|2(x+1)^3| < 1 \iff -1 - 2^{-1/3} < x < -1 + 2^{-1/3}.$$
</s>