I have this issue about some sums:

$\displaystyle \sum_{k=1}^{\infty} \frac{1}{3^k-2^k}$

It must be divergent because $\displaystyle 3^k>2^k$ but I cant show that, any idéas?

Ohh yeah I tried with $\displaystyle 3^k=2^{k*ln(3)}$ but that didn´t really lead me anywhere,

And the second one is $\displaystyle \sum_{k=1}^{\infty}$$\displaystyle \frac{3k^2+3k+1}{k^3(k+1)^3}$

I get it to converge to 0 because the denominator goes towards infinity so much faster, but the books answer is 1. Dont know how to get that.

Any ideas what I am doing wrong?