Question

Is this convergence or divergence?

Answer 1

By the ratio test, the series will converge if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\dfrac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}\right|$

You have

$\displaystyle\lim_{n\to\infty}\left|\dfrac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}\right|=|x|\lim_{n\to\infty}\frac1{n+1}=0<1$

so indeed the series converges.

Furthermore, recalling that

$e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$

it follows that the given series is equivalent to

$\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!}=\sum_{n=0}^\infty\frac{2^n}{n!}-\dfrac{2^0}{0!}=e^2-1$