$\sum \frac{a_n}{\ln a_n}$ converges $\implies \sum \frac{a_n}{\ln (1+n)}$ converges 
Let $0<a_n<1$ be real numbers. Prove that $\sum \frac{a_n}{\ln a_n}$ converges $\implies \sum \frac{a_n}{\ln (1+n)}$ converges 

Note that $\displaystyle \frac{a_n}{\ln (1+n)} \leq \frac{a_n}{-\ln a_n}\iff a_n\geq \frac{1}{n+1}$
If $a_n\geq \frac{1}{n+1}$ for sufficiently many $n$, the series $\sum \frac{a_n}{\ln (1+n)}$ can thefore be compared with $\sum \frac{a_n}{-\ln a_n}$ and we're done. Otherwise, $a_n< \frac{1}{n+1}$ and we're tempted to consider the series $\displaystyle \sum \frac{1}{(n+1)\ln (1+n)}$ which is unfortunately divergent.
That's as I far as I can go with  this problem. Summation by part is a dead end as well.

Great start. 
Now imagine you could compare against $\sum \frac 1 {(n+1)^2\ln(n+1)}$... in your approach, compare the terms of the "new" series against 2 times the terms of the "old" series (times -1). This will lead you to the convergent series for the "remaining" terms.
As you put things together, to make the "sufficiently many $n$" rigorous, compare the new series against the SUM of 2 times the old series (times -1) AND the convergent series $\sum \frac 1 {(n+1)^2\ln(n+1)}$. 
