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|((x^(n+1))/((n+1)!))/((x^n)/(n!))\right|`

You have


`\displaystyle\lim_(n\to\infty)\left|((x^(n+1))/((n+1)!))/((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(x^n)/(n!)`

it follows that the given series is equivalent to


`\displaystyle\sum_(n=1)^\infty(2^n)/(n!)=\sum_(n=0)^\infty(2^n)/(n!)-(2^0)/(0!)=e^2-1`