Prove the convergence of : $\sum \ln(n)/n^{3/2}$ I've been having some issues with what to compare it to. I have a hunch it converges. But I just cannot figure out what I can compare it too. Please help. :)
 A: Here is another way.
You could also use the fact that, for a decreasing sequence $(a_n)$, the sum $ \sum a_n $ converges if and only if $ \sum (2^n a_{2^n}) $ converges.
This is called the Cauchy Condensation Test.
From this, you get rid of the log, since convergence of $ \sum {\log(n) \over n^{3/2}} $ is then equivalent to convergence of
$$ \sum (2^n * \log(2^n)* (2^n)^{-3/2}) = \log(2) \sum {n \over 2^{n/2}} $$
and you can probably assume that the latter converges, or at least hopefully you can show it. (If not, then give me a shout!)
A: Note that for sufficiently large $n$, $\ln(n) \leq n^{1/3}$ and the series
$$
\sum \frac{n^{1/3}}{n^{3/2}} = \sum \frac{1}{n^{7/6}}
$$
converges by the $p$-test since $7/6 > 1$. Thus, the series
$$
\sum \frac{\ln(n)}{n^{3/2}}
$$
converges by the comparison test.
A: Hint: A useful heuristic is that for every $\alpha > 0$, 
$$\ln n < n^{\alpha}$$
eventually. In particular, one might choose a very small value of $\alpha$, so small that
$$\sum_{n = 1}^{\infty} \frac{n^{\alpha}}{n^{3/2}}$$
converges by studying a $p$-series.
