I saw this question here and I tried it on my own. I don't have a background in Probability theory, so I tried to employ algebra only to evaluate this limit.
$$\lim_{n\to \infty}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}$$
My Attempt:
$$\begin{aligned}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}&=\frac{1}{5^n}\sum_{k=0}^{4n} {k+n-1\choose n-1} \left(\frac{4}{5}\right)^{k}\\ &=\frac{1}{5^n}\left(\text{coeff. of }x^{n-1} \text{ in }\sum_{k=0}^{4n}(1+x)^{k+n-1}\left(\frac{4}{5}\right)^k\right)\\ &=\frac{1}{5^n}\left[\text{coeff. of }x^{n-1} \text{ in } \left((1+x)^{n-1}\frac{\left(\frac{4}{5}(1+x)\right)^{4n+1}-1}{\left(\frac{4}{5}(1+x)\right)-1}\right)\right]\end{aligned}$$
I don't know how to proceed because computing the coefficient in the above term is a really long task. Any hints are appreciated. Thanks
$$\begin{aligned}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}&=\frac{1}{5^n}\sum_{k=0}^{4n} {k+n-1\choose n-1} \left(\frac{4}{5}\right)^{k}\\ &=\frac{1}{5^n}\left(\text{coeff. of }x^{n-1} \text{ in }\sum_{k=0}^{4n}(1+x)^{k+n-1}\left(\frac{4}{5}\right)^k\right)\end{aligned}$$
therefore what you re after is
=$\frac{1}{5^n} \sum_{k=0}^{4n}{n-1\choose k+n-1}\left(\frac{4}{5}\right)^k $