TITLE: Calculating limit of a sum
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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: