Sum $\sum_{n=1}^{\infty}[\frac{x(x+n)}{n}]^n$ Consider series $\sum_{n=1}^{\infty}[\frac{x(x+n)}{n}]^n$. Find all values of $x$ such that the series is convergent. 
I'm thinking about using the geometric series. to make the series convergent I need to have 
$$|\frac{x(x+n)}{n}|<1\:\:\text{for all }\:\:n\in \mathbb{N}.$$
Hence I think $|x|<1$, but actually I'm looking for the exact sum of the series. I appreciate any hint.  
 A: If you could find a uniform bound looking like
$$\left|\frac{x(x + n)}{n}\right| \le r < 1$$
then you'd be set, after comparing to a geometric series. Notice that
$$\left|\frac{x(x +n)}{n}\right| \le |x|\left(1 + \frac{|x|}{n}\right).$$
Now provided that $|x| < 1$, choose $n$ large enough that this quantity is strictly less than $1$, for example something like $n > x^2 / (1 - |x|)$. All subsequent terms can be compared with a geometric series, and there are only finitely many preceeding terms. Hence the series is convergent for $|x| < 1$. Using a similar technique, you can study what happens when $|x| > 1$.
If $x = 1$, it is clearly divergent. If $x = -1$, we have
$$\sum_{n = 1}^{\infty} \left(\frac{n - 1}{n}\right)^n$$
and the terms tend to $e^{-1}$, not zero.
A: Cauchy criterion: the positive terms series converges if
$\lim_{n\to \infty } \, \sqrt[n]{a_n}<1$
In our series
$\lim_{n\to \infty } \, \left|\dfrac{x (n+x)}{n}\right|=|x|$
so the series is convergent if $|x|<1$ and is divergent for $|x|>1$
There is no formula for the sum of the series in the general case. 
Hope this helps
Edit
The series for $x=1$ becomes $\sum _{n=1}^{\infty } \left(\frac{n+1}{n}\right)^n$ which clearly diverges because the general term $a_n\to e$ doesn't tend to zero
for $x=-1$ the series becomes $\sum _{n=1}^{\infty }(-1)^n \left(\frac{n-1}{n}\right)^n$ 
this is an alternating series which according to the Leibniz criterion  diverges
$\lim_{n\to \infty } \, \left(\dfrac{n-1}{n}\right)^n=\frac{1}{e}\ne 0$
A: I don't think that there is a closed form, but you can derive some asymptotics.
The general term is $x^n\left(1+\dfrac xn\right)^n$ where the second factor can be approximated by
$$e^x-\frac12 e^xx^2\frac1n+\frac1{24}e^xx^3(3x+8)\frac1{n^2}-\cdots$$
After summation this gives
$$\frac{xe^x}{1-x}+\frac12 e^xx^2\log(1-x)+\frac1{24}e^xx^3(3x+8)\text{Li}_2(x)+\cdots$$
