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I want to make sure my solution is correct, also possibly see some other solution from the people of Math.SE

$$\sum_{n=1}^\infty \frac{\log{n}}{n^c},c\in\mathbb{R}$$

For $c\leq 0$ series clearly diverge, as the sequence does not converge to 0.

For $c\in (0,1]$ I can use the inequality $\frac{1}{n^c}<\frac{\log n}{n^c}$ which holds for almost all $n$ to show that the series diverge by the comparisom test.

For $c> 1$ I used the inequality $\log n < n^\epsilon, \epsilon>0$, once again true for almost all $n$. $\sum \frac{n^\epsilon}{n^c}$ converges for $c>1$ and by comparison, so does the original series.

Is this correct? Thank you for any help.

Dahn
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  • This is about 95% correct, but you need to be careful - you need to pick a good specific epsilon (as a function of $c$) for the $c\gt 1$ case (and note that you say $c\geq 1$ at the start of that paragraph, which is incorrect). For instance, if you have $c=\frac54$ and try to take $\epsilon=\frac13$, that won't work (prove this!). – Steven Stadnicki Jan 18 '14 at 08:50
  • I knew there's a catch somewhere, thanks :) The sign was a typo, corrected. – Dahn Jan 18 '14 at 08:53
  • How about $\epsilon=1-c$ then? – Dahn Jan 18 '14 at 08:55
  • That doesn't work when $c\gt 1$ - you wind up with $\epsilon\lt 0$. (And taking $\epsilon = c-1$ doesn't work either; you end up with the harmonic series, which doesn't converge, while you're trying to show convergence.) – Steven Stadnicki Jan 18 '14 at 09:03
  • Okay, too hasty. Looking at the last series, I need $\epsilon-c<-1$, so $\epsilon<c-1$ should hopefully do the trick.. right? – Dahn Jan 18 '14 at 09:13
  • Yep - but again, you need to keep $\epsilon$ positive. You should be able to get away with just saying 'take $0\lt\epsilon\lt c-1$', but generating an explicit value to use is helpful. – Steven Stadnicki Jan 18 '14 at 09:18
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    Not that it matters, but this sum is $-\zeta'(c)$. – Lucian Jan 18 '14 at 09:59

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