$$\sum_{n=1}^{\infty} \frac{(-2)^{n}}{n^{n}}$$
=
$$\sum_{n=1}^{\infty} \left( \frac{-2}{n}\right)^{n}$$
=
Root test

$$\frac{-2}{n} $$ Doesn't really work :(

The Ratio test, on the other hand, leaves me with
$$\frac{(-2)(n^{n})}{(n+1)^{n+1}}$$

Question

$$\sum_{n=1}^{\infty} \frac{(-2)^{n}}{n^{n}}$$
=
$$\sum_{n=1}^{\infty} \left( \frac{-2}{n}\right)^{n}$$
=
Root test

$$\frac{-2}{n} $$  Doesn't really work :(

The Ratio test, on the other hand, leaves me with 
$$\frac{(-2)(n^{n})}{(n+1)^{n+1}}$$

turingtest · Answer

continue with the ratio test I say

turingtest · Answer

$$\frac{(-2)(n^{n})}{(n+1)^{n+1}}=-2{n^n\over n^{n+1}+O(n)^k}$$where$$k\le n$$

turingtest · Answer

so divide top and bottom by $n^n$

anonymous · Answer

$$ (-2) \left( \frac{n}{n+1} \right)^{n}$$

anonymous · Answer

sorry just had to write that thought down...anyhow

anonymous · Answer

becomes 1/n

anonymous · Answer

so zero

anonymous · Answer

Agreed?

turingtest · Answer

yes, sorry, my power went out :/