<s>[INST] A non-probabilistic take on $\lim_{n\to \infty }\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}$ 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 
 [/INST] $$\begin{aligned}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^n\left(\frac{4}{5}\right)^{k-n}&=\frac{1}{5^n}\sum_{k=0}^{4n} {k+n-1\choose n-1} \left(\frac{4}{5}\right)^{k}\\ &=\frac{1}{5^n}\left(\text{coeff. of }x^{n-1} \text{ in }\sum_{k=0}^{4n}(1+x)^{k+n-1}\left(\frac{4}{5}\right)^k\right)\end{aligned}$$
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 $
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