Show that if an > 0 and Σan is convergent, then Σln(1+an ) is convergent.
limit comparison test with $\displaystyle a_n,$ then let's compute $\displaystyle \lim_{n\to\infty}\frac{\ln\big(1+a_n\big)}{a_n},$ substitute $\displaystyle t=a_n$ then $\displaystyle t\to0$ since $\displaystyle a_n\to0$ because of the convergence of the first series, thus $\displaystyle \lim_{t\to0}\frac{\ln(1+t)}t=1$ and the second series converges.
or for each $\displaystyle n\ge1$ we have $\displaystyle \ln(1+n)\le n$ then $\displaystyle \ln(1+a_n)\le a_n$ and the rest follows by direct comparison test. (Comparison test does apply since $\displaystyle a_n>0,$ so our work is okay.)