$limlimits_{n rightarrow infty} e^{-2n}sum_{k=0}^n frac{(2n)^k}{k!}$

It is possible to calculate the limit $$limlimits_{n rightarrow infty} e^{-2n}sum_{k=0}^n frac{(2n)^k}{k!}$$ 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 $ninmathbb{N}$.

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

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

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

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 $ninmathbb{N}$.

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

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

Since $;lim_limits{nrightarrowinfty}frac{a_{n+1}}{a_n}=frac{2}{e}<frac{4}{5}$, it follows that there exists $n_0inmathbb{N}$ such that $frac{a_{n+1}}{a_n}<frac{4}{5}$ for any $nge n_0$, therefore $frac{a_n}{a_{n-1}}<frac{4}{5}$ for any $nge n_0+1$ and

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

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 $nge n_0+1$,

and by applying Squeeze Theorem we get

$$lim_limits{nrightarrowinfty} frac{S_n}{T_n}=0.$$

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

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

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

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

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Remarks.

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

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