# Convergence of a serie depending on an exponent

For which $k>0$ is $\sum \limits_{n=1}^\infty (\frac {1}{n+n^k})$ convergent.
I tried to prove convergence with the Ration test but it hasn't really worked.
by the P-test and comparison test I found that:
$\sum \limits_{n=1}^\infty (\frac {1}{n+n^k}) < \sum \limits_{n=1}^\infty (\frac {1}{n^k})$
So that the majorant $\frac {1}{n^k}$ is convergent for k>1, therefore the condition for the convergence of the serie is fullfilled if k>1.
Is that correct?

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The idea is correct, but the solution is incomplete - you want to determine what happens for $0 <k < 1$. There may be some $k < 1$ for which the series still converges. But using the $p$-test and comparison test like you did here will give you the answer to that question easily. – Gyu Eun Lee Jan 11 '13 at 8:01
I know that the majorant is in this intervall 0<k<1 divergent, is that explanation enough? – phil Jan 11 '13 at 8:06
Well, that depends on what your teacher wants of you. If he wants to see a proof for everything then probably not. But the methods you've used so far will work for $0 < k \leq 1$ with some easy modifications. – Gyu Eun Lee Jan 11 '13 at 8:10
Yes he wants to see a proof, but I thought the fact that I have proved the p-test for another exercice,would make it possible, that I could use this.Do you have a suggestion for an other proof?Perhaps the integral test? – phil Jan 11 '13 at 8:12
I do, but judging by your work so far I think you're more than capable of doing it by yourself. Just use the comparison test; since you're trying to prove divergence now, find a divergent series smaller than this series in question for $0 < k \leq 1$. – Gyu Eun Lee Jan 11 '13 at 8:15