$$\lim_{n\rightarrow \infty }\sum_{k=n}^{5n}\binom{k-1}{n-1}\left( \frac{1}{5} \right )^n\left( \frac{4}{5} \right )^{k-n}=\text{?}$$
I tried to use formulas $$\sum_{p=0}^k\binom{m+p}m = \binom{m+k+1}{m+1}$$ and $$n!= \sqrt{2\pi n}\left( \frac{n}{e} \right)^n$$ but without any success.
Let $\left[x^j\right]f(x)$ denote the $x^j$ coefficient in $f(x)$. You seek$$\frac15\lim_{n\to\infty}\left(\left[x^{n-1}\right]\sum_{k=n}^{5n}\left(\frac{4+x}{5}\right)^{k-1}\right)=\lim_{n\to\infty}\left[x^{n-1}\right]\frac{\left(\frac{4+x}{5}\right)^{n-1}-\left(\frac{4+x}{5}\right)^{5n}}{1-x}\\=\lim_{n\to\infty}\sum_{j=0}^{n-1}\left[x^j\right]\left(\left(\frac{4+x}{5}\right)^{n-1}-\left(\frac{4+x}{5}\right)^{5n}\right)=1-\lim_{n\to\infty}P(X<n|X\sim\text{Binom}(5n,\,\tfrac15))\\=1-\lim_{n\to\infty}P(X<n|X\sim N(n,\,\tfrac{4n}{5}))=\tfrac12.$$