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Find the value of $\lim\limits_{n \rightarrow \infty} e^{-2n}\sum_{k=0}^n \frac{(2n)^k}{k!}$

A similar type of question was discussed here Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$

I have tried and my answer is $1$

Can we solve it without using Gamma function?

Learning
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2 Answers2

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By using the fact that $k \mapsto \frac{(2n)^k}{k!}$ is increasing for $k \leq 2n$, we have

$$0 \leq e^{-2n}\sum_{k=0}^{n} \frac{(2n)^k}{k!} \leq e^{-2n} (n+1) \frac{(2n)^n}{n!}, $$

Writing $M_n$ for the upper bound in the above inequality and noting that

$$ \frac{M_{n+1}}{M_n} = \frac{e^{-2(n+1)} (n+2) \frac{(2n+2)^{n+1}}{(n+1)!}}{e^{-2n} (n+1) \frac{(2n)^n}{n!}} = 2e^{-2}\frac{n+2}{n+1}\left(1+\frac{1}{n}\right)^n \xrightarrow{n\to\infty} 2e^{-1} < 1, $$

we obtain $M_n \to 0$ and therefore the desired limit is zero. $\square$

Remarks.

  1. Stirling's appoximation immediately tells that $M_n \sim \sqrt{\frac{n}{2\pi}}\left(\frac{2}{e}\right)^n$ as $n\to\infty$.

  2. In general, the CLT tells that, for $\lambda, \mu > 0$, $$ \lim_{n\to\infty} e^{-\lambda n} \sum_{0 \leq k \leq \mu n} \frac{(\lambda n)^n}{k!} = \begin{cases} 0, & \text{if $\mu < \lambda$}, \\ \frac{1}{2}, & \text{if $\mu = \lambda$}, \\ 1, & \text{if $\mu > \lambda$}. \end{cases} $$

Sangchul Lee
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    I wonder why my answer got downvoted. If this is because my previous answer used Stirling's approximation, I now replaced that part with a more elementary proof. – Sangchul Lee Jul 21 '20 at 18:25
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It is possible to calculate the limit $$\lim\limits_{n \rightarrow \infty} e^{-2n}\sum_{k=0}^n \frac{(2n)^k}{k!}$$ without using CLT nor Stirling’s approximation.

Let $$S_n=1+2n+\frac{(2n)^2}{2!}+\frac{(2n)^3}{3!}+\ldots+\frac{(2n)^n}{n!}$$ and $$T_n=\frac{(2n)^{n+1}}{(n+1)!}+\frac{(2n)^{n+2}}{(n+2)!}+\frac{(2n)^{n+3}}{(n+3)!}+\ldots+\frac{(2n)^{2n}}{(2n)!}+\ldots$$ for any $n\in\mathbb{N}$.

So we get that $$e^{2n}=S_n+T_n$$ for any $n\in\mathbb{N}$.

Since the sequence $\left\{\frac{(2n)^k}{k!}\right\}_{k\in\mathbb{N}\cup\{0\}}$ is increasing for $k\le n$ (actually it is increasing for $k\le 2n-1$), it follows that $\;S_n<\frac{(n+1)(2n)^n}{n!}$ for any $n\in\mathbb{N}$.

Moreover, $\;T_n>\frac{(2n)^{2n}}{(2n)!}\;$ for any $\;n\in\mathbb{N}$.

Hence $$0<\frac{S_n}{T_n}<\frac{(n+1)(2n)^n(2n)!}{n!(2n)^{2n}}=\frac{(n+1)(2n)!}{n!(2n)^n}$$ for any $n\in\mathbb{N}$.

Let $\;\;a_n=\frac{(n+1)(2n)!}{n!(2n)^n}\;$ for any $n\in\mathbb{N}$.

It follows that $$\frac{a_{n+1}}{a_n}=\frac{(n+2)(2n+2)!}{(n+1)!(2n+2)^{n+1}}\cdot\frac{n!(2n)^n}{(n+1)(2n)!}=\frac{(n+2)(2n+1)}{(n+1)^2\left(1+\frac{1}{n}\right)^n}$$ for any $n\in\mathbb{N}$.

Since $\;\lim_\limits{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\frac{2}{e}<\frac{4}{5}$, it follows that there exists $n_0\in\mathbb{N}$ such that $\frac{a_{n+1}}{a_n}<\frac{4}{5}$ for any $n\ge n_0$, therefore $\frac{a_n}{a_{n-1}}<\frac{4}{5}$ for any $n\ge n_0+1$ and

$a_n=\frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdot\frac{a_{n-2}}{a_{n-3}}\cdot\ldots\cdot\frac{a_{n_0+1}}{a_{n_0}}\cdot a_{n_0}<a_{n_0}\left(\frac{4}{5}\right)^{n-n_0}$ for any $n\ge n_0+1$.

Hence $$0<\frac{S_n}{T_n}<\frac{(n+1)(2n)!}{n!(2n)^n}=a_n<a_{n_0}\left(\frac{4}{5}\right)^{n-n_0}$$ for any $n\ge n_0+1$,

and by applying Squeeze Theorem we get

$$\lim_\limits{n\rightarrow\infty} \frac{S_n}{T_n}=0.$$

Moreover, $$\lim\limits_{n \rightarrow \infty} e^{-2n}\sum_{k=0}^n \frac{(2n)^k}{k!}=\lim\limits_{n\rightarrow\infty}\frac{S_n}{e^{2n}}=\lim\limits_{n\rightarrow\infty}\frac{S_n}{S_n+T_n}=\lim\limits_{n\rightarrow\infty}\frac{\frac{S_n}{T_n}}{\frac{S_n}{T_n}+1}=0$$

Angelo
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