Considering the convergence of intergral: $${\Large I=}\int_1^{+\infty} \frac{1}{\ln^2(x+1)}dx$$
Using Dirichlet theorem, we have the answer. Or $\displaystyle{0<\frac1x<\frac{1}{\ln^2(x+1)}}$, hence $I$ diverge.
But I think there will have a simplier answer. Please help.
Another line of thoughts that leads to similar observations: The map $x \mapsto \frac{1}{\log^{2}(x+1)}: [1, +\infty[ \to \mathbb{R}$ is decreasing. Note that $k > \log^{2}(k+1)$ for large $k$; this implies $$ \frac{1}{\log^{2}(k+1)} > \frac{1}{k} $$ for large $k$; so by comparison test the series $\sum_{k} \frac{1}{\log^{2}(k+1)}$ diverges. By integral test we then see the divergence of the improper integral.
