# Is this series convergent or divergent ? $\sum\limits_{n=1}^\infty \frac{(-1)^n \ln(n)}{n}$ [duplicate]

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

How do I determine if this converges or diverges? I should be able to show this with just the basic library of Calc II tests (Geometric, p-series, Divergence (nth term) test, Integral test, Direct Comparison test, Alternating Series Test, or Ratio test) None of these involved derivatives, either

## marked as duplicate by heropup, Semiclassical, user147263, Jack's wasted life, ShaileshMar 25 '16 at 0:10

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• You should also realize that users can see your question history, and can track whether or not you are (re)posting duplicate questions, which is against site policy. Again: (1) don't do it. (2) it's pointless, anyway. – heropup Mar 24 '16 at 21:32

## 2 Answers

Try the integral test on the absolute series. You will get that it diverges. Then, using the alternating series test, you will get that the series is conditionally convergent.

$\displaystyle \int_{1}^{\infty} \frac{\ln (x)}{x} dx = \displaystyle \frac{1}{2}\left( \lim_{t\to \infty} \ln ^2|t| - \ln ^2(1)\right) = \infty$

Then, since $\frac{\ln(n)}{n}$ is monotonically decreasing for all $n\geq 2$, $\displaystyle \lim_{n\to \infty} \frac{\ln(n)}{n} =0$, and the absolute series diverges, the series is conditionally convergent.

• Note that since $(\ln n)/n > 1/n$ for $n \geq 3$, you could also prove divergence of the absolute series via comparison to the harmonic series. – Michael Seifert Mar 24 '16 at 21:30
• I took the integral of the absolute series, the absolute value of $(-1)^x$ is just 1. – Sorey Mar 24 '16 at 21:38
• To show a series converges conditionally it must diverge absolutely – Sorey Mar 24 '16 at 21:41
• Ok, so your integral test was to show the absolute series diverged? I did that using direct comparison against harmonic. Was simpler. So, I am now just interested in how to show the original series converges. You just used Alt. Series test for that part? But, the orig. series is not decreasing. 0,.35,.37,.35, etc – JackOfAll Mar 24 '16 at 21:43
• Well you can use the comparison test as well. There's often multiple ways to show series diverge. The original series will converge conditionally cause of reasons stated above in my answer – Sorey Mar 24 '16 at 21:45

It's convergent, and we have:

$$\sum_{n = 1}^{+\infty} (-1)^n \frac{\ln(n)}{n} = \frac{1}{2}\left(2\gamma\ln(2) - \ln^2(2)\right) \approx 0.159869$$

Where $\gamma$ is the Euler-Mascheroni constant.

Hint: Alternating Term Series Test.

• How did you calculate the value? I've never seen this one before... – zz20s Mar 24 '16 at 21:36