Convergence of $\sum_{n=1}^{\infty} \sqrt{\frac{\ln(n+1)}{1+\ln(n+1)}}\left(\sqrt{\frac{1+\ln(n+1)}{1+\ln\left(n\right)}}-1\right).$

Question

I came across the sum $\sum_{n=1}^{\infty} a_n$ while computing an upper bound, where $$a_n = \sqrt{\frac{\ln\left(n+1\right)}{1+\ln\left(n+1\right)}}\left(\sqrt{\frac{1+\ln\left(n+1\right)}{1+\ln\left(n\right)}}-1\right).$$
I am struggling to decide whether this sum converges. Graphically the summand goes to $0$ as $n \to \infty$, but the sum seems to grow extremely slowly, like $\sqrt{\ln(\ln(n))}$, but all of this is just a hunch from the graphs I have plotted. I am not sure how to prove whether this sum diverges or not. I tried comparison tests with $b_n = \frac{1}{n^{\alpha}}$, $\alpha = \frac{7}{6}, \frac{5}{4}$, but they failed for large values of $n$.

Answer 1

Note that $$ \sqrt {\frac{{1 + \log (n + 1)}}{{1 + \log (n)}}} - 1 = \sqrt {1 + \frac{{\log (n + 1) - \log n}}{{1 + \log (n)}}} - 1 \sim \sqrt {1 + \frac{1}{{n\log (n)}}} - 1 \sim \frac{1}{{2n\log (n)}} $$ and $$ \sqrt {\frac{{\log (n + 1)}}{{1 + \log (n + 1)}}} \sim 1 $$ as $n\to +\infty$. Thus, $$ a_n \sim \frac{1}{{2n\log (n)}} $$ as $n\to +\infty$, whence the series diverges by the limit comparison test (since $\sum \frac{1}{{n\log (n)}}$ diverges by the integral test or the Cauchy condensation test).