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

The above series is the one I want to prove convergent. I took $\frac{\ln n}{n}$ but I didn't find anything that I could do with it. I tried to compare it with another series, didn't find a good one. Any idea for this?

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    I changed "ln \ n" to "\ln n". The backslash not only prevents italicization, but also results in proper spacing between "ln" and "n". Generally with \det, \max, \sup, \sin, etc., the backslash causes the typesetting conventions appropriate to the situation to be followed. For example "\max_{x\in A}" in a "displayed" setting gives you this $\displaystyle\max_{x\in A}$, and in an "inline" setting gives you $\max_{x\in A}$.2012-01-10

1 Answers 1

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Hint: Try using the alternating series test.

Remark: If you are interested in more then convergence, we can evaluate this explicitly and show that $\sum_{n=1}^\infty (-1)^n \frac{\log n}{n}=\gamma \log 2 -\frac{(\log 2)^2}{2}$ where $\gamma$ is the Euler-Mascheroni constant. More generally, we can show that for any $k$ $\sum_{n=1}^{\infty}\frac{(-1)^{n}\log^{k}n}{n}=\frac{(-1)^{n}\left(\log2\right)^{k+1}}{k+1}+(-1)^{k-1}\sum_{j=0}^{k-1}\gamma_{j}\binom{k}{j}\log^{k-j}2,$ where the $\gamma_i$ are the Stieltjes constants. See this answer, or this post for specific details regarding these last two facts. (The links are nearly identical)

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    @NikhilBellarykar: Also take a look at the divergence test, also sometimes called the the term test: http://en.wikipedia.org/wiki/Term_test2012-01-10