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Evaluate the limit $$ \lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(\sum_{k=0}^{n}{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}} \right)} $$

I use $$e^{n}=1+n+\frac{n^{2}}{2!}+\cdots+\frac{n^{n}}{n!}+\frac{1}{n!}\int_{0}^{n}{e^{x}(n-x)^{n}dx}$$ but I don't know how to evaluate $$ \lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(e^{n}-2\frac{1}{n!}\int_{0}^{n}{e^{x}(n-x)^{n}dx} \right) }$$

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I guess the second part converges to zero, and for the first part i would use stirling, but that would mean it diverges – Dominic Michaelis Mar 8 '13 at 8:09
The answer is $\frac{4}{3}$, which we can prove by utilizing rlgordonma's calculation. – Sangchul Lee Mar 8 '13 at 8:27
up vote 3 down vote accepted

In this answer, it is shown, using integration by parts, that $$ \sum_{k=0}^n\frac{n^k}{k!}=\frac{e^n}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\tag{1} $$ Subtracting both sides from $e^n$ gives $$ \sum_{k=n+1}^\infty\frac{n^k}{k!}=\frac{e^n}{n!}\int_0^n e^{-t}\,t^n\,\mathrm{d}t\tag{2} $$ Substtuting $t=n(s+1)$ and $u^2/2=s-\log(1+s)$ gives us $$ \begin{align} \Gamma(n+1) &=\int_0^\infty t^n\,e^{-t}\,\mathrm{d}t\\ &=n^{n+1}e^{-n}\int_{-1}^\infty e^{-n(s-\log(1+s))}\,\mathrm{d}s\\ &=n^{n+1}e^{-n}\int_{-\infty}^\infty e^{-nu^2/2}\,s'\,\mathrm{d}u\tag{3} \end{align} $$ and $$ \begin{align} \Gamma(n+1,n) &=\int_n^\infty t^n\,e^{-t}\,\mathrm{d}t\\ &=n^{n+1}e^{-n}\int_0^\infty e^{-n(s-\log(1+s))}\,\mathrm{d}s\\ &=n^{n+1}e^{-n}\int_0^\infty e^{-nu^2/2}\,s'\,\mathrm{d}u\tag{4} \end{align} $$ Computing the series for $s'$ in terms of $u$ gives $$ s'=1+\frac23u+\frac1{12}u^2-\frac2{135}u^3+\frac1{864}u^4+\frac1{2835}u^5-\frac{139}{777600}u^6+O(u^7)\tag{5} $$ In the integral for $\Gamma(n+1)$, the odd powers of $u$ in $(5)$ are cancelled and the even powers of $u$ are integrated over twice the domain as in the integral for $\Gamma(n+1,n)$. Thus, $$ \begin{align} 2\Gamma(n+1,n)-\Gamma(n+1) &=\int_n^\infty t^n\,e^{-t}\,\mathrm{d}t-\int_0^n t^n\,e^{-t}\,\mathrm{d}t\\ &=n^{n+1}e^{-n}\int_0^\infty e^{-nu^2/2}\,2\,\mathrm{odd}(s')\,\mathrm{d}u\\ &=n^{n+1}e^{-n}\left(\frac4{3n}-\frac8{135n^2}+\frac{16}{2835n^3}+O\left(\frac1{n^4}\right)\right)\\ &=n^ne^{-n}\left(\frac43-\frac8{135n}+\frac{16}{2835n^2}+O\left(\frac1{n^3}\right)\right)\tag{6} \end{align} $$ Therefore, combining $(1)$, $(2)$, and $(6)$, we get $$ \begin{align} \frac{n!}{n^n}\left(\sum_{k=0}^n\frac{n^k}{k!}-\sum_{k=n+1}^\infty\frac{n^k}{k!}\right) &=\frac{e^n}{n^n}\left(\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t-\int_0^n e^{-t}\,t^n\,\mathrm{d}t\right)\\ &=\frac43-\frac{8}{135n}+\frac{16}{2835n^2}+O\left(\frac1{n^3}\right)\tag{7} \end{align} $$

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We can write that integral as

$$\int_0^n dx \: e^x (n-x)^n = n^{n+1} \int_0^1 du \: e^{n u} (1-u)^n = n^{n+1}\int_0^1 du \: e^{n [u+\log(1-u)]} $$

Now, as $n \rightarrow \infty$, that last integral is dominated by contributions near $u=0$. We may then use the first term in the Taylor expansion of the exponential, and the integral is, to lowest order of approximation

$$n^{n+1} \int_0^{\infty} du e^{-n u^2/2} = n^{n+1} \sqrt{\frac{2 \pi}{n}} $$

Use Stirling's approximation to get the rest of the story.


It was pointed out that we will need to go to the next order of approximation. In this case, use $u+\log{(1-u)} \approx -u^2/2-u^3/3$. We then get, as an approximation to the integral

$$n^{n+1} \int_0^{\infty} du e^{-n u^2/2} e^{-n u^3/3}$$

Note that, form the lowest order of approximation, we are only considering $u \sim 1/\sqrt{n}$. This means that $n u^3 \sim 1/\sqrt{n}$ and we can Taylor expand this exponential for small argument to get the approximation

$$n^{n+1} \int_0^{\infty} du\: e^{-n u^2/2} \left (1-\frac{n u^3}{3}\right) = n^{n+1} \left (\sqrt{\frac{2 \pi}{n}} - \frac{2}{3 n} \right )$$

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This approach will work, but it's necessary to go beyond the lowest order of approximation for the integral and use both the $u^2/2$ and $u^3/3$ terms in the series for $\log (1-u)$. – David Moews Mar 8 '13 at 9:02
@DavidMoews: fortunately, that is straightforward, as I will show above. – Ron Gordon Mar 8 '13 at 9:15

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