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

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

Attempt


*

*Let's denote $F$ as 
$$
\begin{align*}
& \lim_{n\to\infty}\sum_{k=n}^{5n}{k-1 \choose n-1}\left(\frac{1}{5}\right)^{n}\left(\frac{4}{5}\right)^{k-n} = \\ 
&= \lim_{n\to\infty}\left(\frac{1}{5}\right)^{n}\frac{1}{(n-1)!}\sum_{k=n}^{5n}\left(\frac{4}{5}\right)^{k-n}\frac{(k-1)!}{(k-n)!} =  \\ &= \lim_{n\to\infty}\left(\frac{1}{5}\right)^{n}\frac{1}{(n-1)!}\sum_{k=0}^{4n}\left(\frac{4}{5}\right)^{k}\frac{(k+n-1)!}{k!} = F
\end{align*}
$$

*Try to establish a lower and upper bounds of denoted function $F$, lower bound set to 0, and upper bound to $$\lim_{n\to\infty}\left(\frac{1}{5}\right)^{n}\frac{1}{(n-1)!}\sum_{k=0}^{\infty}\left(\frac{4}{5}\right)^{k}\frac{(k+n-1)!}{k!}$$
The sum inside of a limit equals to $(1 + \frac{4}{5})^{n}$ (by using Taylor series). By substituting this into the obtained upper bound, we're getting $0$ as an answer.

 A: 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 
$$\lim_{n \to \infty} \sum_{k=n}^{5n} \binom{k-1}{n-1} \left(\frac{1}{5}\right)^n \left(\frac{4}{5}\right)^{k-n} = \frac{1}{2}.$$
A: 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.
A: 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 $\Pr(Z\le 0) = \dfrac 1 2.$
