Calculate the limit including binomial coefficients $\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$
It's clear that we can simplify the limit a little bit, after which we get:
$\lim\limits_{n\to\infty}{(\frac{1}{4})^{n}\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{4}{5})^{k}}$
I could further simplify the expression, but I feel like there's a more elegant solution. 
Give me a hint, please
 A: Hint for a probabilistic proof: look at the negative binomial distribution as the sum of independent geometric distributions, and apply the central limit theorem.
A: Let $\left[x^j\right]f(x)$ denote the $x^j$ coefficient in $f(x)$. You seek$$\frac15\lim_{n\to\infty}\left(\left[x^{n-1}\right]\sum_{k=n}^{5n}\left(\frac{4+x}{5}\right)^{k-1}\right)=\lim_{n\to\infty}\left[x^{n-1}\right]\frac{\left(\frac{4+x}{5}\right)^{n-1}-\left(\frac{4+x}{5}\right)^{5n}}{1-x}\\=\lim_{n\to\infty}\sum_{j=0}^{n-1}\left[x^j\right]\left(\left(\frac{4+x}{5}\right)^{n-1}-\left(\frac{4+x}{5}\right)^{5n}\right)=1-\lim_{n\to\infty}P(X<n|X\sim\text{Binom}(5n,\,\tfrac15))\\=1-\lim_{n\to\infty}P(X<n|X\sim N(n,\,\tfrac{4n}{5}))=\tfrac12.$$
A: Using identify ${n\choose r} = \frac{n}{r}\cdot {{n-1} \choose {r-1}}$
$$\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$$
= $$ \lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n} \frac{n}{k}{k \choose n}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}} $$.
Now, Divide both sides in the expansion of $(1+x)^n=nC0+nC1 x+...+nCn x^n$ by x and integrate both sides from 0 to 1. 
Substitute the identity in the limit expression and evaluate it.
