4
$\begingroup$

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

Improve this question
$\endgroup$
  • $\begingroup$ I had a typo in the original post, where in the first display $a_n $ should have been $n $ - fixed now $\endgroup$ – Richard Aug 31 '18 at 15:32
  • $\begingroup$ Is $a_n > 0$ for all $n$? $\endgroup$ – Daniel Fischer♦ Aug 31 '18 at 15:36
  • $\begingroup$ @DanielFischer Yes $\endgroup$ – Richard Aug 31 '18 at 15:39
  • $\begingroup$ Perhaps it was one of these questions that you saw. $\endgroup$ – Antonio Vargas Aug 31 '18 at 18:21
  • $\begingroup$ Why does this not tend to 1? Do you have the original proof? $\endgroup$ – Henry Lee Aug 31 '18 at 18:53
3
$\begingroup$

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 $$\sum_{k = 0}^{n} \frac{a_n^k}{k!} \sim e^{a_n}\,.$$

Improve this answer
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – user Aug 31 '18 at 15:54
  • $\begingroup$ @gimusi It depends on $n$, but not on $k$. $\endgroup$ – Daniel Fischer♦ Aug 31 '18 at 15:56
  • $\begingroup$ But isn't it equivalent to assume $e^{a_n} = \sum_{k = 0}^{\infty} \frac{a_n^k}{k!}$? $\endgroup$ – user Aug 31 '18 at 16:00
  • 1
    $\begingroup$ 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.) $\endgroup$ – Daniel Fischer♦ Aug 31 '18 at 16:03
  • $\begingroup$ Were you perhaps thinking about $$\sum_{k = 0}^{n} \frac{a_k^k}{k!}\,,$$ @gimusi? $\endgroup$ – Daniel Fischer♦ Aug 31 '18 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.