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I wonder how to calculate the following limit: $$ \lim_{n\rightarrow\infty}\frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}} $$ In the first sight, I think it should be zero, because exponential function is much faster than polynomial. But the upper of the expression is the Maclaurin polynomial of $e^{n}$. With the growth of n, it approaches to $e^{n}$. Consider there is a roughly way to estimate the remainder of $e^{n}$ rather than $$ R_{n+1}(n)=\frac{\xi^{n+1}}{(n+1)!} \ for \ \ \xi\in(n,+\infty) $$ Because $$ \frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}}=1-\frac{R_{n+1}(n)}{e^{n}} $$ But it's hard to continue.

jintok
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  • Playing around with Wolfram: The partial sum formula for $e^n$ is $\frac{e^n \Gamma(n+1, n)}{\Gamma(n+1)}$. So as $n \to \infty$, your limit seems to approach $\frac{1}{2}$. – Andrey Kaipov Sep 26 '14 at 06:49
  • With regard to your first comment: "an exponential grows faster than a polynomial" means that $e^n$ grows faster than $p(n)$, where $p$ is a polynomial of fixed degree. This is not the case in your example. – David Sep 26 '14 at 06:52
  • There are numerous interesting solutions here, http://math.stackexchange.com/questions/160248/evaluating-lim-n-to-infty-e-n-sum-limits-k-0n-fracnkk . – Travis Willse Sep 26 '14 at 06:53
  • thanks all above, seems i give up too early. – jintok Sep 26 '14 at 07:11

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Let a be the difference error between $e^n$ and maclaurins series the value of the expression now becomes $1-\frac a{e^n}$ and now as n tends to $\infty $ the difference error a tends to zero and $e^n$ tends to $\infty$ and therefore the limit tends to 1 which implies that they both grow at same rate.

Jasser
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