0
$\begingroup$

$$\sum^\infty_{n=0} \frac{e^n}{n!}$$ I am pretty sure already it converges, I'm trying to find to what. $$\sum^\infty_{n=0} \frac{e^n}{n!}=\sum^\infty_{n=1}\prod^n_{j=1}\frac{e}{j}$$ $$\sum^\infty_{n=0}\frac{\sum^\infty_{j=0}\frac{n^j}{j!}}{n!}=\sum^\infty_{n=0}\sum^\infty_{j=0}\frac{n^j}{n!j!}=\sum^\infty_{j=0}\frac{1}{j!}\sum^\infty_{n=0}\frac{n^j}{n!}$$ from here, i have no idea.

$\endgroup$
3
  • 1
    $\begingroup$ what's the taylor expansion of $e^x$? $\endgroup$ – user58955 Nov 17 '20 at 7:32
  • $\begingroup$ I'll try. thanks for the help. $\endgroup$ – razivo Nov 17 '20 at 7:33
  • $\begingroup$ Are you trying to obtain a proof for the expansion of $\exp x$? If yes, then tell me. If no, then remember it; the series you're talking about is the expansion of $e^x$. $\endgroup$ – ultralegend5385 Nov 17 '20 at 7:41
4
$\begingroup$

We will use the Taylor series fromula: $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$ This formula converges to the value of the function if and only if the function is analytic in an open disk centered at 0.
Now, let $f(x)=e^x$. We know that $$\frac{d^n}{dx^n}e^x=e^x\quad \forall n\ge0$$ so that $$\frac{d^n}{dx^n}e^x\bigg\rvert_{0}=e^0=1$$ Substituting this in the original Taylor series formula yields \begin{align} e^x&=\sum_{n=0}^{\infty}\frac{\frac{d^n}{dx^n}e^x\big\rvert_{0}}{n!}x^n\\ &=\sum_{n=0}^{\infty}\frac{x^n}{n!} \end{align} Substituting $x=e$ gives $$\sum_{n=0}^{\infty}\frac{e^n}{n!}=e^e\approx15.142622$$

$\endgroup$
5
  • $\begingroup$ Can you give the assumptions with which the formula "$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$" holds ? It is definitely not true for every function $f$, the series may diverge or not be equal to $f(x)$. $\endgroup$ – TheSilverDoe Nov 17 '20 at 7:53
  • $\begingroup$ @TheSilverDoe did it. Please check if I am right. $\endgroup$ – Leonhard Euler Nov 17 '20 at 7:58
  • $\begingroup$ Thank you for the edit, but this is still not right :) Even if the radius of convergence is $+\infty$, the series can be not equal to the function... Consider $f = \exp(-x)$ if $x > 0$ and $f(x)=0$ if $x \leq 0$. $\endgroup$ – TheSilverDoe Nov 17 '20 at 8:01
  • $\begingroup$ @TheSilverDoe sorry, I didn't realize it. Now I updated it and I am almost sure that it's true. $\endgroup$ – Leonhard Euler Nov 17 '20 at 8:08
  • $\begingroup$ Ok, now this is correct ! Even if one should ask why $\exp$ is analytic ;) But I upvote your answer, thanks for takinf the time of correcting. $\endgroup$ – TheSilverDoe Nov 17 '20 at 8:12

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.