the sum of a series I am stuck on the computation of the following sum:
$$\sum_{k=0}^{\infty} {\Big( {\frac{q}{k+1}} \Big)}^k ,$$
where $k$ is a natural number, and $0<q<1$.
 A: Let's modify the proof of the Sophomore's Dream:
Start by applying the change of variables $x\mapsto e^{-x}$:
$$
\begin{align}
\int_0^1(-qx\log(x))^k\,\mathrm{d}x
&=\int_\infty^0(qxe^{-x})^k\,\mathrm{d}e^{-x}\\
&=q^k\int_0^\infty x^ke^{-(k+1)x}\,\mathrm{d}x\\
&=\frac{q^k}{(k+1)^{k+1}}\int_0^\infty x^ke^{-x}\,\mathrm{d}x\\
&=\frac{q^kk!}{(k+1)^{k+1}}\tag{1}
\end{align}
$$
Plugging $(1)$ into $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ yields
$$
\begin{align}
\sum_{k=0}^\infty\left(\frac{q}{k+1}\right)^{k+1}
&=\int_0^1qx^{-qx}\,\mathrm{d}x\\
&=\int_0^q(x/q)^{-x}\,\mathrm{d}x\tag{2}
\end{align}
$$
Take the derivative of $(2)$ with respect to $q$:
$$
\begin{align}
\sum_{k=0}^\infty\left(\frac{q}{k+1}\right)^k
&=1+\int_0^q(x/q)^{1-x}\,\mathrm{d}x\\
&=1+q\int_0^1x^{1-qx}\,\mathrm{d}x\tag{3}
\end{align}
$$
I don't think there is a simplification beyond this.
A: I think the converge of the series is very obvious. It is non-negative, so lower bounded trivially by zero. Also $\frac{q}{k+1} < q$, hence,
$$\sum_{k=0}^n \left(\frac{q}{k+1} \right)^k < \sum_{k=0}^n q^k $$
The RHS converges to $1/(1-q)$. Hence the sequence converges. But it is a very loose upper bound.
A: I have no idea to express the partial sum with $n$ and $q$.
However, I am curious about that if the complete problem is ask you whether it's convergence or not? If so, I can tell you the answer is sure.
In the series, the $k$-th term is $(\frac{q}{k+1})^{k}$, denoted it by $a_{k}(>0)$, then , the quotient of the adjacent two terms is 
$$\frac{a_{k+1}}{a_{k}}=\frac{q^{k+1}(k+1)^{k}}{q^{k}(k+2)^{k+1}}<q<1$$
by criterion of series, it's convergence.
