Q:

How to prove this series converges $\sum_{n=2}^\infty \frac{\ln(n)}{n^{3/2}}$?

What test do you use to prove that
$$\sum_{n=2}^\infty \frac{\ln(n)}{n^{3/2}}$$
converges? 
I tried the limit comparison test using $\frac{1}{n^{3/2}}$ as the comparison, but it did not converge.

A:

Hint: You can use integral test.
Added: You need to consider the integral

$$ \int_{2}^{\infty} \frac{\ln(x)}{x^{3/2}}dx ,  $$

which can be integrated using integration by parts.

A:

Apply Cauchy Condensation Test.
Let $a_n =  \dfrac{\ln(n)}{n^{3/2}}$, then $a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}}$
Therefore, 
$$a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}} = \frac{n\ln 2 }{2^{3n/2}} \leq \ln2\frac{n}{2^n}$$
Thus 
$$ \sum_{n=2}^{\infty}\dfrac{\ln 2^n }{(2^n)^{3/2}}  \leq \sum_{n=2}^{\infty} \ln2\frac{n}{2^n}.$$
The RHS is a convergent series, by the Ratio Test, as you should verify.

A:

You can use that $\ln n$ goes to infinitely much more slowly that any (positive) power of $n$, that is: for every $\alpha>0$, $\ln n < n^\alpha$ for $n$ large enough. You can apply this to $\alpha=1/4$ says (any $\alpha$ strictly between $0$ and $1/2$ will do, so let's say $1/4$), so $\ln n < n^{1/4}$ for $n$ large enough, hence $\ln n / n^{3/2} < 1/n^{5/4}$ for $n$ large enough, and since the series $\sum 1/n^{5/4}$ converges (as does $\sum 1/n^\beta$ for any $\beta>1$), you're done.

