# Convergence of $\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p}$ [duplicate]

Consider the series

$$\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p}$$

where $p>0$. For what $p$, the series converges?

This question appears in the Tier 1 Analysis exam (2014/08) of Indiana University Bloomington. And I modify it a little to make it look more tight...

$$\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p} \ge \sum_{n=2}^{\infty} \frac{1}{n(\log n)^p},$$ which diverges if $p\le 1$ (well-known and not hard to show with the standard techniques).
Further, $$\sum_{n=2}^{\infty} \frac{1}{(\log 1)^p+(\log 2)^p+\cdots +(\log n)^p} \le \sum_{n=2}^{\infty} \frac{1}{n/2 \cdot (\log (n/2))^p},$$ which converges if $p> 1.$
• Your claim is false. Try any $p>1$ and you will see... – Simply Beautiful Art Aug 1 '17 at 20:05
• Ah, okay, you fixed it to $p<1$ $\ddot\smile$ – Simply Beautiful Art Aug 1 '17 at 20:15
• The second inequality is not obvious to me when $n$ is odd. Maybe split the series into two parts, one for even $n$ and one for odd $n$? – edm Aug 4 '17 at 3:55