Series(IIII...)

Calculus Level 2

$\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } } = ?$

A) ${ e }^{ x }$

B) $0$

C) ${ e }^{ { x }^{ 2 } }$

D) $\infty$

Note: $x\neq 0$

A B C D

1 solution

Munem Shahriar
Nov 11, 2017

By the ratio test , it converges to $e^x.$

Ratio test:

If there exists an $N$ so that for all $n \geq N,$ $a_n \ne 0$ and $\lim_{n \to \infty} \left|\dfrac{a_n + 1}{a_n}\right| = L$

Here $L = 0,$ . And 0 is less than 1. Hence $\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } } = e^x$

Note: $\left|\dfrac{a_n + 1}{a_n}\right| = \left|\dfrac{\dfrac {x^{(n+1)}}{(n+1)!}}{\dfrac {x^n}{n!}}\right|$

Can you be more specific? I think testing out here would help other solvers. Thanks.

A Former Brilliant Member - 3 years, 7 months ago

Is that okay?

Munem Shahriar - 3 years, 7 months ago