Alternating series: $\sum\limits_{n= 1}^{\infty} (-1)^{n-1} \frac{\ln(n)}{n}$ convergence?

Question

Determine whether the series converges absolutely, conditionally or diverges?

$$\sum\limits_{n= 1}^{\infty} (-1)^{n-1} \frac{\ln(n)}{n}$$

I know that $\sum|a_{n}|$ diverges by using the comparison test:

$$\frac{\ln(n)}{n} > \frac{1}{n}$$ and the smaller, r.h.s being the divergent harmonic series.

So, should my conclusion for the alternating series be divergent or convergent conditionally^*?

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^{* How to estimate whether the alternating series terms are cancelling?}

Answer 1

Let $f(x)=\frac{\ln x}{x}$ so $f'(x)=\frac{1-\ln x}{x^2}\le 0$ for $x\ge e$ and so the sequence $\left(\frac{\ln n}{n}\right)_{n\ge3}$ is decreasing to $0$. Apply now the alternating series criteria to conclude the convergence of the series.

Answer 2

Since $\frac{\log n}{n}$ is a decreasing function on $n\geq 3$ and $\{(-1)^n\}_{n\geq 1}$ is a sequence with bounded partial sums, the series is convergent by Dirichlet's test. Moreover, by exploiting the integral representation for $\log(n)$ provided by Frullani's theorem, we have:

$$ S=\sum_{n\geq 1}\frac{(-1)^{n-1}\log(n)}{n}=\int_{0}^{+\infty}\frac{e^{-x}\log(2)-\log(1+e^{-x})}{x}\,dx\tag{1}$$ and by the integral representation for the Euler-Mascheroni constant we have: $$ \sum_{n\geq 1}\frac{(-1)^{n-1}\log(n)}{n} = \color{red}{\frac{\log^2(2)}{2}-\gamma\log(2)}\approx -0.1598689 .\tag{2}$$

Answer 3

It converges conditionally because of alternating series test.

You have

$$\lim_{n\to\infty}(-1)^n\frac{\ln n}n=0$$

and for $n$ large enough

$$\left\vert (-1)^n\frac{\ln n}n\right\vert\geq\left\vert (-1)^{n+1}\frac{\ln (n+1)}{n+1}\right\vert.$$