Question

Determine whether the following series converges or diverges.

Answer 1

have you tried anything yet ?

Answer 2

I was advised to do the ratio test, im doing that right now.

Answer 3

just for a variety, here is one way to work the sum :

\[e^x = \sum\limits_{k=0}^{\infty} \dfrac{x^n}{n!} \]

differentiate and get
\[e^x = \sum\limits_{k=0}^{\infty} \dfrac{nx^{n-1}}{n!} = \]

multiplying \(x\) both sides gives
\[xe^x = \sum\limits_{k=0}^{\infty} \dfrac{x^{n}}{(n-1)!} \]

Answer 4

differentiate again two times and playing with a bit, you can find the sum..

Answer 5

i have a typo, let k = n above. .

Answer 6

with \(a_k\) as your general term,
\[
\frac{a_{k+1}}{a_k}
= \frac{\frac{2^{k+2}(k+1)^2}{k!}}{\frac{2^{k+1}k^2}{(k-1)!}}
= \frac{2}{1} \frac{(k+1)^2}{k^2}\frac1k.
\]
Right?

Answer 7

@Albertoimus

Answer 8

I haven't do the math, but I am almost sure that this converges since factorial grows much faster than the numerator. \(n!\sim\sqrt{2\pi n}\,n^ne^{-n}\).

Answer 9

@reemii