the sum of a series

Question

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 $0<q<1$.

Answer 1

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.

Answer 2

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.

Answer 3

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.