# does the infinite series $\sum^{\infty}_{n=1} (-1)^n \frac {\log(n)}n$ converge?

does the infinite series $\sum^{\infty}_{n=1} (-1)^n \frac {\log(n)}n$ converge?

For this one I tried absolute convergence then I applied the integral test but I realized that $\log^2(x)/2$ does not converge so I know that that won't work. Any help? Also I know the limit of $a_n$ as n approaches $\infty=0$ however I am not sure if it is non decreasing

• Show that (after a while) $\frac{\log n}{n}$ is decreasing. So, after a while, we have an alternating series. The after a while decreasing can be done by differentiating $\frac{\log x}{x}$. – André Nicolas Mar 25 '14 at 8:30
• Yes, the point is that it is much easier to test for the convergence of an alternating series, as long as the individual terms in the sum tend to zero in absolute value, and are eventually decreasing in absolute value. The convergence need not be absolute. – Geoff Robinson Mar 25 '14 at 8:33
• Leibniz criterion it is called. – user76568 Mar 25 '14 at 8:38

The sequence $\log n\over n$ is decreasing for $n>2$ because the function $x\mapsto{\log x\over x}$ is decreasing in $(e,\infty)$: $$f'(x)={1-\log x\over x^2}.$$
The convergence might be established using the alternating series test since $\frac{\log n}n$ is a monotonically decreasing sequence for large $n$.
We have that $$f(x)=\frac{\log x}x$$ and $$f'(x)=\frac{1-\log x}{x^2}$$ is negative for $x>e$. So the sequence is monotonically decreasing for $n\ge 3$.
Alternating series converge if the absolute value of the sequence converges monotonically to zero. In this case, $n>\log n$ for large $n$, so everything is fine.