Find limit $\lim_{n\to\infty}\sum_{k=n}^{5n}{k-1\choose n-1}\left(\frac{1}{5}\right)^{n}\left(\frac{4}{5}\right)^{k-n}$

A probabilistic interpretation:

Let $X_1, \dots, X_{5n}$ be independent Bernoulli variables each with mean $p = 1/5$. Note that when the sum $X := \sum X_i$ is at least $n$, we can define the index $T$ of the $n$-th variable with $X_i = 1$ (i.e. we let $T$ be the $n$-th smallest element of the set $\{i : X_i = 1\}$). When $X < n$, we can set $T = \bot$ to indicate that $T$ is undefined.

Now note that for $n \leq k \leq 5n$, we have $T = k$ if and only if $X_k = 1$ and exactly $n-1$ of the $k-1$ variables $X_1, \dots, X_{k-1}$ have $X_i = 1$. This means $$\mathbb{P}[T = k] = \binom{k-1}{n-1} \left(\frac{1}{5}\right)^n \left(\frac{4}{5}\right)^{k-n}$$ hence $$\mathbb{P}[X \geq n] = \mathbb{P}[T \neq \bot] = \sum_{k=n}^{5n} \mathbb{P}[T = k] = \sum_{k=n}^{5n} \binom{k-1}{n-1} \left(\frac{1}{5}\right)^n \left(\frac{4}{5}\right)^{k-n}.$$

But by the central limit theorem, we have $$\lim_{n \to \infty} \mathbb{P}[X \geq n] = \lim_{n \to \infty} \mathbb{P} \left[ \frac{X - n}{\sqrt{5n \mathrm{Var}[X_1]}} \geq 0 \right] = \mathbb{P}[Y \geq 0]$$ for $Y$ a standard normal variable (i.e. $Y \sim N(0, 1)$). This probability is just $1/2$ (the standard normal distribution is symmetric), hence $$\lim_{n \to \infty} \sum_{k=n}^{5n} \binom{k-1}{n-1} \left(\frac{1}{5}\right)^n \left(\frac{4}{5}\right)^{k-n} = \frac{1}{2}.$$

We can rewrite the sum as $$ \eqalign{ & S(n) = \sum\limits_{k = n}^{5n} {\left( \matrix{ k - 1 \cr n - 1 \cr} \right)p^{\,n} q^{\,k - n} } \quad \left| {\,1 \le n} \right.\quad = \cr & = p^{\,n} \sum\limits_{k = n}^{5n} {\left( \matrix{ k - 1 \cr k - n \cr} \right)q^{\,k - n} } = \sum\limits_{k = 0}^{4n} {\left( \matrix{ k + n - 1 \cr k \cr} \right)\left( {1 - q} \right)^{\,n} q^{\,k} } \cr} $$ where as usual we put $q=1-p$.

The summand is $$ {\left( \matrix{ k + n - 1 \cr k \cr} \right)\left( {1 - q} \right)^{\,n} q^{\,k} } $$ which is the pmf of the Negative Binomial distribution $NB(k;\,n,q)$.

Thus our sum is the CDF of the above distribution computed at $4n$, which is $$ \eqalign{ & S(n) = \sum\limits_{k = 0}^{4n} {\left( \matrix{ k + n - 1 \cr k \cr} \right)\left( {1 - q} \right)^{\,n} q^{\,k} } = 1 - I_{\,q} (4n + 1,n) = \cr & = I_{\,p} (n,4n + 1) = {{{\rm B}\left( {p;\;n,4n + 1} \right)} \over {{\rm B}\left( {n,4n + 1} \right)}} \cr} $$ where $I_x$ is the Regularized Incomplete Beta function.

The mean and variance of the NB distribution are $$ \mu = {{qn} \over {1 - q}}\quad \sigma ^{\,2} = {{qn} \over {\left( {1 - q} \right)^{\,2} }} $$ and for large $n$ it will converge to the Normal distribution with standard variable $$ {{x - \mu } \over {\sigma \sqrt 2 }} = {{\left( {1 - q} \right)} \over {\sqrt {2qn} }}\left( {x - {{qn} \over {1 - q}}} \right) $$

Concerning the asymptotics for $n \to \infty$, the NB will converge to the Normal distribution and therefore $S(n)$ will converge to the CDF of the Normal distribution for $x=4n$ $$ \eqalign{ & S(n) = \sum\limits_{k = 0}^{4n} {\left( \matrix{ k + n - 1 \cr k \cr} \right)\left( {1 - q} \right)^{\,n} q^{\,k} } \approx \cr & \approx \Phi \left( {{{\left( {1 - q} \right)} \over {\sqrt {2qn} }}\left( {4n - {{qn} \over {1 - q}}} \right)} \right) = \cr & = \Phi \left( {{{\left( {1 - q} \right)\sqrt n } \over {\sqrt {2q} }}\left( {4 - {q \over {1 - q}}} \right)} \right) \approx \cr & \approx H\left( {4 - {q \over {1 - q}}} \right) \cr} $$ where $H$ is the Heaviside step function , with $$H(0)=1/2$$, which is the case in your question.

Imagine an infinite sequence of independent trials with probability $1/5$ if success on each trial.

On average, then it takes $5$ trials to get one success.

Let $X$ be the number of trials needed to get $n$ successes.

Then the expected value of $X$ is $5n.$

The variance of $X$ is $20n.$

That last is somewhat more work to prove, but it is $n$ times the variance of the number of trials needed to get one success, since $X$ is the sum of $n$ independent copies of that random variable.

Thus the random variable $\dfrac{X-5n}{\sqrt{20n}}$ has expected value $0$ and standard deviation $1.$ And as $n\to\infty,$ the distribution of this random variable approaches that of the standard normal random variable $Z.$

$$ \frac{n-5n}{\sqrt{20n}} \le \frac{X-5n}{\sqrt{20n}} \le \frac{5n-5n}{\sqrt{20n}} = 0. $$ So the limit is $\Pr(Z\le 0) = \dfrac 1 2.$