Question

How would one go about finding the convergence or divergence of the series `n^2*(-4)^(n+1)/((n+2)!)` ?

The series starts from n = 1 to `oo`

Answer 1

This series is convergent. The easiest way to prove this would be by using the ratio test
Wikipedia link

Explanation:

Borrowing the notation of the wikipedia article, we have

`a_n =n^2*(-4)^(n+1)/((n+2)!)`

Thus

`L = lim_{n to infty} | a_{n+1} /a_n |`
`= lim_{n to infty} |(n+1)^2*(-4)^(n+2)/((n+3)!) {(n+2)!}/{n^2 (-4)^{n+2}}|`
`= lim_{n to infty} |{-4(n+1)^2}/{n^2(n+3)}| = 0`

Since `L=0<1` the series is convergent.