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I recall a question where it was proved that $$\lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=\frac12.$$This seems to remain true when $n$ is replaced by $n+c $ where $c$ is some constant, while apparently, if $a_n=o(n)$, $$\sum_{k=0}^n\frac{a_n^k}{k!}\sim e^{a_n}.$$How to prove it? Maybe the method used in the related question is also useful here, but I couldn't find it

Richard
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1 Answers1

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For every $0 < \varepsilon < e^{-1}$, choose $N(\varepsilon)$ such that $a_n < \varepsilon n$ for $n \geqslant N(\varepsilon)$. Then we have \begin{align} e^{a_n} - \sum_{k = 0}^{n} \frac{a_n^k}{k!} &= \sum_{k = n+1}^{\infty} \frac{a_n^k}{k!} \\ &< \sum_{k = n+1}^{\infty} \frac{(\varepsilon n)^k}{k!} \\ &< \varepsilon^{n+1}\sum_{k = n+1}^{\infty} \frac{n^k}{k!} \\ &< \varepsilon^{n+1} e^n \\ &< \varepsilon \end{align} for $n \geqslant N(\varepsilon)$. So we even have $$e^{a_n} - \sum_{k = 0}^{n} \frac{a_n^k}{k!} \to 0\,,$$ not only $$\sum_{k = 0}^{n} \frac{a_n^k}{k!} \sim e^{a_n}\,.$$

Daniel Fischer
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  • How is the step $e^{a_n} - \sum_{k=0}^{n} \frac{a_n^k}{k!}= \sum_{k=n+1}^{\infty} \frac{a_n^k}{k!}$ motivated taking into account that $a_n$ is varying? – user Aug 31 '18 at 15:54
  • @gimusi It depends on $n$, but not on $k$. – Daniel Fischer Aug 31 '18 at 15:56
  • But isn't it equivalent to assume $e^{a_n} = \sum_{k = 0}^{\infty} \frac{a_n^k}{k!}$? – user Aug 31 '18 at 16:00
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    But that's a given. $$e^x = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$ is the definition of $e^x$. (Yes, it is possible to define it differently, then one needs to prove the power series expansion, but that's standard.) – Daniel Fischer Aug 31 '18 at 16:03
  • Were you perhaps thinking about $$\sum_{k = 0}^{n} \frac{a_k^k}{k!},,$$ @gimusi? – Daniel Fischer Aug 31 '18 at 16:06
  • Yes of course, it wasn't clear at first that we are fixing $n$ and $a_n$. It is more clear now. Thanks – user Aug 31 '18 at 16:06