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

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*?

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

• en.wikipedia.org/wiki/Alternating_series_test Sep 11 '16 at 17:11
• The Euler-Maclaurin Formula shows \begin{align} \sum_{n=1}^{2N}\frac{(-1)^{n-1}\log(n)}{n}&=-\log(2)\sum_{n=1}^N \frac1n +\sum_{n=N+1}^{2N}\frac{\log(n)}{n}\\\\ &=-\log(2)\left(\log(N)+\gamma +O\left(\frac1N\right)\right)+\int_N^{2N}\frac{\log(x)}{x}\,dx+O\left(\frac{\log(N)}{N}\right)\\\\ &=-\log(2)\left(\log(N)+\gamma +O\left(\frac1N\right)\right)\\\\ &+\frac12 \left(\log^2(2N)-\log^2(N)\right)\\\\ &+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2)-\gamma \log(2)+O\left(\frac{\log(N)}{N}\right) \end{align} Now let $N\to \infty$. Jun 11 '20 at 16:55

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.

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: $$\sum_{n\geq 1}\frac{(-1)^{n-1}\log(n)}{n} = \color{red}{\frac{\log^2(2)}{2}-\gamma\log(2)}\approx -0.1598689 .\tag{2}$$

• Which integral representation are you alluding to, exactly? The evaluation of that integral doesn't seem as simple as you are making it out to be (at least to one who is less experienced with those sorts of things, such as myself). Feb 6 '18 at 0:56
• Another way ...The Euler-Maclaurin Formula shows \begin{align} \sum_{n=1}^{2N}\frac{(-1)^{n-1}\log(n)}{n}&=-\log(2)\sum_{n=1}^N \frac1n +\sum_{n=N+1}^{2N}\frac{\log(n)}{n}\\\\ &=-\log(2)\left(\log(N)+\gamma +O\left(\frac1N\right)\right)+\int_N^{2N}\frac{\log(x)}{x}\,dx+O\left(\frac{\log(N)}{N}\right)\\\\ &=-\log(2)\left(\log(N)+\gamma +O\left(\frac1N\right)\right)\\\\ &+\frac12 \left(\log^2(2N)-\log^2(N)\right)\\\\ &+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2)-\gamma \log(2)+O\left(\frac{\log(N)}{N}\right) \end{align} Now let $N\to \infty$. Jun 11 '20 at 16:56

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

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