Does the series $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n(\ln n)^2}$ absolutely converge?
The index n=1 kept me hanging too. This is an item I saw in one of my books. Isn't the series wrong as the first term of the series will have a 0 in the denominator?
Yes, it does converge (use the alternating series test). It also converges absolutely since $$ \sum_{n=3}^\infty\frac1{n\log^2n}\le\int_2^\infty\frac1{x\log^2x}\mathrm dx=\frac1{\log2}. $$