Calculating limit of a sum

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 $0.91413582462$

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: