Conclude if the given series converges or diverges and justify the conclusion

Test the following series for convergence or divergence. Justify the decision.

$\sum_{n=2}^{\infty} \frac{\log n}{n \sqrt{n+1}}.$

The series converges. To show this, first we note that

$\frac{\log n}{n \sqrt{n+1}} \leq \frac{\log n}{n^{\frac{3}{2}}}.$

Thus, if we can show that the series

$\sum_{n=2}^{\infty} \frac{\log n}{n^{\frac{3}{2}}}$

converges then the series in question converges by the comparison test. To show that this series converges we apply the limit comparison test with

$a_n = \frac{\log n}{n^{\frac{3}{2}}}, \qquad b_n = \frac{n^{\varepsilon}}{n^{\frac{3}{2}}} \qquad 0 < \varepsilon < \frac{1}{2}.$

We know that $\sum b_n$ converges since it is of the form $\sum \frac{1}{n^s}$ for $s > 1$ . Furthermore,

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\log n}{n^{\varepsilon}} = 0$

for $\varepsilon > 0$ . Thus,

$\sum_{n=2}^{\infty} \frac{\log n}{n^{\frac{3}{2}}}$

converges which implies the convergence of

$\sum_{n=2}^{\infty} \frac{\log n}{n \sqrt{n+1}}.$

Stumbling Robot