Calculating limit of a sum Helllo everyone,
I have to calculate a particular limit that contais a sum and I have no idea how to solve such problem. The task is to calculate this limit: 
$$\lim_{n\to \infty}\left(\frac n6\sum_{i=0}^\infty \left(\frac 56\right)^i\left(1-\left(\frac56\right)^i\right)^{n-1}\right) $$
I will be grateful for any hints or solutions. 
 A: Some playing around, but not an answer.  If we set $x=\left(\frac 56\right)^i$ we can find the largest term in the sum by differentiating $x(1-x)^{n-1}$ and setting to zero.  It turns out the maximum is at $x=\frac 1n$ or $i=\frac {\log n}{\log 1.2}$.  The value of the maximum term is about $\frac 1{ne}$.  If we plot the terms of the sum they are sharply peaked around the maximum and the width appears indpendent of $n$.  The figure below is for $n=10000$, where the peak is at $i=50$ or $51$. The horizontal axis is $i$ and the vertical axis is the term in the sum 

I also plotted $n=10, 100, 1000$ and the width and shape of the peak do not seem to change.  I wrote a Python program to sum the series out to $i=100000$, summing from the top down so we don't lose significance adding the tiny terms.  The result was consistent to better than ten figures for all the $n$ I tried from $10$ to $10^6$ at $0.91413582462$
A: I might have a start. I need to sort out error terms with my approximations. 
I'm getting about $\frac{1}{6 \ln{6/5}}$ but my steps might be wrong. 
Let:
$$\alpha=\frac{5}{6}$$
$$f(n)=\frac{1}{6}\frac{n}{n-1}(n-1)\sum_{i=0}^\infty\alpha^i(1-\alpha^i)^{n-1}$$
So:
$$f(n+1)=\frac{1}{6}\frac{n+1}{n}(n)\sum_{i=0}^\infty\alpha^i(1-\alpha^i)^{n}$$
Here's a questionable step.
Does this hold?: 
$$(1-\alpha^i)^n=(1-n\alpha^i/n)^n=e^{-n\alpha^i}$$
Then for large n:
$$f(n+1)=\frac{1}{6}\sum_{i=0}^\infty \ n\alpha^ie^{-n\alpha^i}$$
Then we approximate the sum with an integral substituting x for i.
$$f(n+1)\approx\frac{1}{6}\int_{0}^\infty na^xe^{-na^x}dx$$
Let $u=n\alpha^x$
Then: 
$du=n \ln{\alpha} \alpha^x dx$
So :
$$f(n+1)\approx \frac{1}{6}\int_{n}^0 \frac{1}{\ln{\alpha}}e^{-u}du$$
Integrating:
$$f(n+1)\approx \frac{1}{6\ln{\alpha}}=\frac{1}{6\ln{\alpha}}(-e^{-u})|_n^0=\frac{-1+e^{-n}}{6\ln{\alpha}}$$
Looks accurate:

A: Also not an answer, but you can replace the infinite sum by a finite one as shown below.  A blunder in my earlier attempt to carry out this calculation was responsible for my erroneous assertion that the sum diverged as $\ n\rightarrow\infty\ $.
\begin{eqnarray}
\sum_{i=0}^\infty \left(\frac 56\right)^i\left(1-\left(\frac56\right)^i\right)^{n-1} &=&
\sum_\limits{i=0}^\infty\left(\frac{5}{6}\right)^i\sum_\limits{j=0}^{n-1}{n-1\choose j}\left(-1\right)^j\left(\frac{5}{6}\right)^{ij}\\
&=&\sum_\limits{j=0}^{n-1}\left(-1\right)^j{n-1\choose j}\sum_\limits{i=0}^\infty\left(\frac{5}{6}\right)^{i\left(j+1\right)}\\
&=& \sum_\limits{j=0}^{n-1}\frac{\left(-1\right)^j{n-1\choose j}}{1-\left(\frac{5}{6}\right)^{j+1}}\\
\end{eqnarray}
