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It is known around here that $$\lim_{n\to\infty }e^{-n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)=\frac12$$ So what about the following limit: $$L=\lim_{n \to \infty}n \left(\frac{1}{2}-\frac{1}{e^n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)\right)$$ My thought was to apply Stolz-Cesaro theorem, after rewritting as $$L=\lim_{n \to \infty}\frac{ \left(\frac{1}{2}-\frac{1}{e^n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)\right)}{\frac1n}$$ would produce: $$L=\lim_{n \to \infty}\frac{e^{-(n+1)}\left(1+(n+1)+\frac{(n+1)^2}{2!}+...+\frac{(n+1)^{n+1}}{(n+1)!}\right)-e^{-n}\left(1+n+\frac{n^2}{2!}+...+\frac{n^n}{n!}\right)}{\frac1{n(n+1)}}$$ $$=\lim_{n \to \infty}\frac{e^{-n}\left(\frac{1-e}{e} +\frac1e\frac{(n+1)^{n+1}}{(n+1)!} +n\left(\frac{\left(1+\frac1n\right)}{e}-1\right) +\frac{n^2}{2!}\left(\frac{\left(1+\frac1n\right)^2}{e}-1\right)+\cdots +\frac{n^n}{n!}\left(\frac{\left(1+\frac1n\right)^n}{e}-1\right)\right)}{\frac1{n(n+1)}}$$ But I dont see how can I go further, can you help me evaluate it? Also, are limits of this form already known?

Zacky
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  • Where is that first equation from? Surely $e^{-1}\cdot\left(1+1\right)\neq \frac12$. Or did you mean this as an asymptotic relation? – Jam Aug 15 '18 at 10:41
  • You mean this one: https://math.stackexchange.com/questions/160248/evaluating-lim-limits-n-to-infty-e-n-sum-limits-k-0n-fracnkk ? – Zacky Aug 15 '18 at 10:44
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    Right, you should probably put the $\lim_{n\to+\infty}$ in there, otherwise it isn't true. – Jam Aug 15 '18 at 10:45
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    From plotting $y=\lfloor x\rfloor\left(\frac12-e^{-\lfloor x \rfloor} \cdot \sum_{i=0}^{\lfloor x\rfloor}\frac{\lfloor x \rfloor ^i}{i!} \right)$ in Desmos, I think the expression is divergent. https://imgur.com/a/t7ENPKv – Jam Aug 15 '18 at 10:52
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    Probably not, but you could potentially still get a nice asymptotic relationship out of it since the purple graph appears to approach a certain curve as $x\to+\infty$. It's often good to graph limits to get a feel for whether they are convergent (although this method isn't necessarily foolproof). – Jam Aug 15 '18 at 10:58
  • In fact, I believe the expression is asymptotic to $\frac{n}{2}-\frac{\Gamma(n+1,n)}{\Gamma(n)}$, where $\Gamma(n+1,n)$ is the upper incomplete gamma function and $\Gamma(n)$ is the standard gamma function. (imgur) (desmos) – Jam Aug 15 '18 at 11:41

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The follow equation is from Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$.

\begin{align} e^{-n}\sum_{k=0}^n\frac{n^k}{k!} &=\frac{1}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\\ &=\frac12+\frac{2/3}{\sqrt{2\pi n}}+O(n^{-1})\tag{11} \end{align} Use this equation we know: $$\lim_{n\to\infty}\sqrt{n} \left(e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}-\frac{1}{2}\right)=\frac{2}{3\sqrt{2\pi }}.$$ So your limit is $\infty$.

Riemann
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