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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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Use the alternating series test. –  Jim Belk Jan 10 '12 at 17:50
    
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 $\max_{x\in A}$. –  Michael Hardy Jan 10 '12 at 19:18

1 Answer 1

up vote 5 down vote accepted

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: I don't exactly understand what you are trying to ask, can you elaborate? –  Eric Naslund Jan 10 '12 at 18:10
    
@NikhilBellarykar: Careful, the converse of the alternating series test is not true, so you cannot use it to tell you if a series diverges. For that you need to do something else. –  Eric Naslund Jan 10 '12 at 18:35
    
This link uses the alternating series to prove divergence of a series on similar lines, so I guess I am right perhaps, but would like your opinion on it. math.ufl.edu/~vatter/teaching/m8w10/m8l09.pdf –  Nikhil Bellarykar Jan 10 '12 at 18:43
    
@NikhilBellarykar: Which problem are you referring to? Example 2 uses the divergence test which is slightly different, if the limit of the terms doesn't go to zero, then it must diverge. –  Eric Naslund Jan 10 '12 at 18:52
    
yes, example 2 says that the limit of terms doesnt go to zero so the series diverges. the limit condition is one of the 2 conditions-other being the terms should be monotonically decreasing, and if any one of them is not satisfied, the series diverges. I now think I havent been able to really show what I claimed to show, so I am deleting the comment. Thank you for your patience. –  Nikhil Bellarykar Jan 10 '12 at 18:58

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