Calculate the limit including binomial coefficients

Question

$\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$

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

Answer 1

Hint for a probabilistic proof: look at the negative binomial distribution as the sum of independent geometric distributions, and apply the central limit theorem.

Answer 2

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.

Answer 3

Let $\left[x^j\right]f(x)$ denote the $x^j$ coefficient in $f(x)$. You seek$$\frac15\lim_{n\to\infty}\left(\left[x^{n-1}\right]\sum_{k=n}^{5n}\left(\frac{4+x}{5}\right)^{k-1}\right)=\lim_{n\to\infty}\left[x^{n-1}\right]\frac{\left(\frac{4+x}{5}\right)^{n-1}-\left(\frac{4+x}{5}\right)^{5n}}{1-x}\\=\lim_{n\to\infty}\sum_{j=0}^{n-1}\left[x^j\right]\left(\left(\frac{4+x}{5}\right)^{n-1}-\left(\frac{4+x}{5}\right)^{5n}\right)=1-\lim_{n\to\infty}P(X<n|X\sim\text{Binom}(5n,\,\tfrac15))\\=1-\lim_{n\to\infty}P(X<n|X\sim N(n,\,\tfrac{4n}{5}))=\tfrac12.$$