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This question already has an answer here:

$\lim_{n\rightarrow\infty}\frac{1+\frac{n}{1!}+\cdot+\frac{n^n}{n!}}{e^n}=\frac12$

Taking the first $n$ terms of the Taylor series of $e^n$ as the numerator, the limit is true or false? How to prove?

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marked as duplicate by Paramanand Singh limits Jan 24 '18 at 6:24

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Assuming that we work $$a_n=e^{-n}\sum_{k=0}^n \frac{n^k}{k!} $$by the definition of the incomplete gamma function $$a_n=\frac{\Gamma (n+1,n)}{n \Gamma (n)}$$ We have the relation $$\Gamma (n+1,n)=n \,\Gamma (n,n)+e^{-n}\, n^n$$ which makes $$a_n=\frac{ n^{n-1}}{e^n\,\Gamma (n)}+\frac{\Gamma (n,n)}{\Gamma (n)}$$ The first term tends to $0$ when $n$ becomes large; to prove it, take its logarithm and use Stirling approximation to get $$\log\left(\frac{ n^{n-1}}{e^n\,\Gamma (n)} \right)=-\frac{1}{2} \log \left({2 \pi n}\right)-\frac{1}{12 n}+O\left(\frac{1}{n^{5/2}}\right)$$

For the second term, if you look here, you will notice the asymptotics $$\Gamma(n,n) \sim n^n e^{-n} \sqrt{\frac{\pi}{2 n}}$$ So, neglecting the first term, we have, for large $n$ $$a_n\sim \frac{ n^n e^{-n} }{\Gamma(n)}\sqrt{\frac{\pi}{2 n}}$$ Take logarithms and use Stirling approximation to get $$\log(a_n)=-\log (2)-\frac{1}{12 n}+O\left(\frac{1}{n^{5/2}}\right)$$ Continue with Taylor $$a_n=e^{\log(a_n)}=\frac{1}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^{2}}\right)$$ If you use the better asymptotics given in the link $$\Gamma(n,n) = n^n e^{-n} \left [ \sqrt{\frac{\pi}{2 n}} - \frac{1}{3 n} + O\left ( \frac{1}{n^{3/2}} \right ) \right ]$$ doing the same, you should end with $$a_n=\frac 12-\frac{1}{3 \sqrt{2 \pi n} }-\frac{1}{24 n}+O\left(\frac{1}{n^{3/2}}\right)$$

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You have $$ \frac {\sum_{k=0}^n\frac {n^k}{k!}}{e^n} = \frac {e^n-\sum_{k=n+1}^\infty \frac {n^k}{k!}}{e^n} = 1-\frac {\sum_{k=n+1}^\infty \frac {n^k}{k!}}{e^n}\to1. $$ To see that the quotient goes to zero, since the tail of the sum is the remainder in the Taylor polynomial of $e^n$ around $0$, we have $$ \sum_{k=n+1}^\infty \frac {n^k}{k!}=\frac{e^{c(n)}n}{(n+1)!}, $$ where $c(n)$ is some number between $0$ and $n$. Then, as $e^{c(n)}\leq e^n$, $$ \frac{\sum_{k=n+1}^\infty \frac {n^k}{k!}}{e^n}=\frac1{e^n}\,\frac{e^{c(n)}n}{(n+1)!}\leq\frac{n}{(n+1)!}\to0. $$

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  • $\begingroup$ Why $\frac {\sum_{k=n+1}^\infty \frac {n^k}{k!}}{e^n}$ is $0$? $\endgroup$ – W. mu Jan 24 '18 at 3:39
  • $\begingroup$ The numerator goes to zero, and the denominator goes to infinity. $\endgroup$ – Martin Argerami Jan 24 '18 at 3:40
  • $\begingroup$ But there are infinite terms.. $\endgroup$ – W. mu Jan 24 '18 at 3:41
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    $\begingroup$ Shouldn't the limit be $1/2$ as asked in question instead of $1$? $\endgroup$ – Paramanand Singh Jan 24 '18 at 6:18
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    $\begingroup$ The remainder in Taylor's theorem should be $e^{c(n)} n^{n+1}/(n+1)!$ $\endgroup$ – Paramanand Singh Jan 24 '18 at 6:20

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