# $\sum^\infty_{n=0} \frac{e^n}{n!}$

$$\sum^\infty_{n=0} \frac{e^n}{n!}$$ 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.

• what's the taylor expansion of $e^x$? – user58955 Nov 17 '20 at 7:32
• I'll try. thanks for the help. – razivo Nov 17 '20 at 7:33
• 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 $e^x$. – ultralegend5385 Nov 17 '20 at 7:41

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 $$\sum_{n=0}^{\infty}\frac{e^n}{n!}=e^e\approx15.142622$$

• 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 $f(x)$. – TheSilverDoe Nov 17 '20 at 7:53
• @TheSilverDoe did it. Please check if I am right. – Leonhard Euler Nov 17 '20 at 7:58
• 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 $x \leq 0$. – TheSilverDoe Nov 17 '20 at 8:01
• @TheSilverDoe sorry, I didn't realize it. Now I updated it and I am almost sure that it's true. – Leonhard Euler Nov 17 '20 at 8:08
• 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. – TheSilverDoe Nov 17 '20 at 8:12