comparison test to show that $\sum_{n=1}^{\infty}\frac{1}{(n+2)\sqrt{ \ln ^3(n+3)}}$ converges

Question

As the title says, I know that this sum converges and I want to find a suitable comparison test.

Cauchy's root test and d'Alembert's ratio test gave inconclusive results. According to wolfram this sum converges according to comparison test but I'm having problems thinking of one.

I thought maybe of $\frac{1}{(n+2)\sqrt{ \ln ^2(n+3)}}=\frac{1}{(n+2)\ln (n+3)}$ but the sum $\sum_{n=1}^{\infty}\frac{1}{(n+2)\ln (n+3)}$ diverges (again, according to wolf).

Anyone has any better ideas how to approach this?

Answer 1

You may use Cauchy condensation test. By Cauchy condensation test, we only check the convergence of $$\sum_{k=0}^\infty 2^k \frac{1}{(2^k+2)\ln^{3/2}(2^k+3)}$$ but $$\sum_{k=1}^\infty 2^k \frac{1}{(2^k+2)\ln^{3/2}(2^k+3)}\le \sum_{k=1}^\infty \frac{1}{\ln^{3/2}(2^k)}=\sum_{k=1}^\infty \frac{1}{k^{3/2}(\ln 2)^{3/2}}.$$

Answer 2

HINT: $$ \sum_{n=2}^{\infty}\frac1{x(\ln x)^a} $$ converges for $a>1$ and diverges for $a\leq1$. It is easy to obtain from integral test.