Sum $\sum_{n=0}^\infty \frac{x^n}{k(n+k)}$ I have to find the sum of the following power series:
$$\sum_{n=0}^\infty \frac{x^n}{k(n+k)}$$ where $k$ is a real number greater than $0$. I tried to get to the $\log(1+x)$ expansion but the $n+k$ at the denominator doesn't let me do so. What can I do to find the sum of the series? Thanks
 A: Let $P(x) = \sum\limits_{n = 0}^\infty \frac{x^n}{k(n + k)}$.
Then we see that $x^{k} P(x) = \sum\limits_{n = 0}^\infty \frac{x^{n + k}}{k(n + k)}$.
Then we see that $\frac{d}{dx} x^k P(x) = \sum\limits_{n = 1}^\infty \frac{x^{n + k - 1}}{k} = \frac{x^k}{k} \sum\limits_{n = 1}^\infty x^{n - 1} = \frac{x^k}{k} \frac{1}{1 - x}$.
Now in general, the quantity $\frac{x^k}{k} \frac{1}{1 - x}$ does not have a nice anti-derivative. However, in the case that $k$ is a known integer, we can explicitly take the antiderivative to get a nice closed form for $P(x)$.
Edit: in fact, whenever $k$ is rational, we can come up with some closed form for the integral. If $k = \frac{p}{q}$, then we can proceed with the substitution $u^q = x$ to get the integral of $\frac{u^p}{k} \frac{1}{1 - u^q} q u^{q - 1}$, which is a rational function and can therefore be integrated using the method of partial fractions.
In general, one must express the sum in terms of the Lerch Transcendent function. But this is essentially just a rephrasing of the original sum and offers little insight.
A: As already said $$f_k(x)=\sum_{n=0}^\infty \frac{x^n}{k(n+k)}=\frac 1k \Phi (x,1,k)$$where appears the Hurwitz-Lerch transcendent function.
If $k$ is an integer, you can write it as
$$f_k(x)=-\frac {x^{1-k}}k \Bigg[\frac{\log (1-x)}{x}+\sum_{n=1}^{k-1}\frac{x^{n-1}} n \Bigg]$$
For the case where $k$ is not an integer, you could be interested by this paper.
Using @Mark Saving's approach, for the most general case we can also write
$$f_k(x)=\frac{x }{k (k+1)}\, _2F_1(1,k+1;k+2;x)+\frac{1}{k^2}$$ where appears the gaussian hypergeometric function and $k$ can be any number.
