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I have to compute

$$ \lim_{n\to\infty} \exp(-n)\left(1+n+\frac{n^2}{2}+\ldots+\frac{n^n}{n!} \right)$$

I think the value is 1, but i don't know how to proof this. Do I have to estimate the remainder of the Taylor expansion of the exponential function? Does anyone have a hint for me?

So do I have to show that $$ R_{n} (x) = \int_{-\infty}^{x} \frac{\exp(t)}{n!} \left(x-t\right)^{n}\mathrm{d}t $$ goes to 0? Or is there an easier way?

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marked as duplicate by Antonio Vargas, hardmath, M Turgeon, Daniel Rust, Thomas Andrews Dec 9 '13 at 18:01

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Do you think Riemann sum will help somehow? –  Praphulla Koushik Dec 9 '13 at 15:31
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Numerics strongly suggest the answer is $\frac{1}{2}$, but I have no proof yet. –  universalset Dec 9 '13 at 15:34

1 Answer 1

If you express this in terms of the series-residual $$ f(n)= \exp(-n)\cdot (\exp(n) - (n^{n+1}/(n+1)! + n^{n+2}/(n+2)! + ... ) $$ you can write $$ f(n) = 1 - {(n^{n+1}/(n+1)! + n^{n+2}/(n+2)! + ... ) \over \exp(n)} $$ and this can furtherly be reduced to $$ f(n) = 1 - {n^n\over n!} \cdot {(n/(n+1) + n^2/(n+1)/(n+2) + ... ) \over \exp(n)} $$ and then to $$ f(n) = 1 - {(n/e)^n\over n!} \cdot \left( {1 \over 1+1/n} + {1 \over (1+1/n)(1+2/n)} + ... \right) $$ Call that infinite sum in the parenthese as $R(n)$, then it seems as if a not bad accurate estimate is $ R(n) \approx 1.174 n^{0.506}$ (taken by some values by nonlinear trendline using Excel, so the values might be improvable). Then - I think- the Stirling-formula for the estimate of the factorial should come into play...

As it was already conjectured in the comments, it seems that the final result goes to 1/2 by simple numerical tests using the series representations of $R(n)$.

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$R(n)$ is probably going to be $\sqrt{\frac{n\pi}{2}}$ (plus some error terms). –  universalset Dec 9 '13 at 17:03

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