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

Question

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

Answer 1

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