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How can we compute the series $\displaystyle \sum_{n\geq 1}\frac{(-1)^n \ln n}{n}$?

I know it is $\eta '(1)$ , where $\eta$ is the $\eta$ Dirichlet Function , i know its value. But I don't know how to compute it.

An approach I tried is to expand the series, then gather together the odd and the even terms , use $\zeta$ (Riemann's function) and that's all. Then no idea.

Any ideas are welcome.

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  • $\begingroup$ I want to compute it... meaning I want to get the result wikipedia says. $\endgroup$ – Tolaso Sep 16 '14 at 18:43
  • $\begingroup$ What does Wikipedia say it is? $\endgroup$ – robjohn Sep 16 '14 at 23:37
  • $\begingroup$ It gives the result $\gamma \ln 2- \dfrac{\ln^2 2}{2}$. $\endgroup$ – Tolaso Sep 17 '14 at 7:32
  • $\begingroup$ Here's a link of sos440's blog article. I think it is quite nice. $\endgroup$ – karvens Sep 27 '14 at 7:24
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This is given in equation $(8)$ of this answer using the Euler-Maclaurin Sum Formula, which says that $$ \sum_{k=1}^n\frac{\log(k)}{k}=C+\frac{\log(n)^2}{2}+O\left(\frac{\log(n)}{n}\right)\tag{1} $$ for some constant $C$, and that $$ \sum_{k=1}^n\frac1{k}=\log(n)+\gamma+O\left(\frac1n\right)\tag{2} $$ Note that $$ \begin{align} \sum_{k=1}^{2n}(-1)^k\frac{\log(k)}{k} &=2\sum_{k=1}^{n}\frac{\log(2k)}{2k} -\sum_{k=1}^{2n}\frac{\log(k)}{k}\\ &=\sum_{k=1}^{n}\frac{\log(k)}{k}+\log(2)\sum_{k=1}^{n}\frac1{k} -\sum_{k=1}^{2n}\frac{\log(k)}{k}\\ &=C+\frac{\log(n)^2}{2}+O\left(\frac{\log(n)}{n}\right)\\ &+\log(2)\left(\log(n)+\gamma+O\left(\frac1n\right)\right)\\ &-C-\frac{\log(2n)^2}{2}-O\left(\frac{\log(2n)}{2n}\right)\\ &=\gamma\log(2)-\frac{\log(2)^2}{2}+O\left(\frac{\log(n)}{n}\right)\tag{3} \end{align} $$ Let $n\to\infty$ in $(3)$ and we get $$ \sum_{k=1}^\infty(-1)^k\frac{\log(k)}{k}=\gamma\log(2)-\frac{\log(2)^2}{2}\tag{4} $$ Note that it doesn't matter what the constant $C$ is, as it gets cancelled out of the calculations.

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  • $\begingroup$ Nice answer! +1. $\endgroup$ – Olivier Oloa Sep 17 '14 at 5:38
  • $\begingroup$ @robjohn very nice answer! $\endgroup$ – Tolaso Sep 17 '14 at 7:33
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Let $s>1$. Observe that $$\displaystyle \sum_{n\geq 1}\frac{(-1)^n }{n^s}=\frac{1}{2^s}\sum_{n\geq 1}\frac{1}{n^s}-\sum_{n\geq 0}\frac{1}{(2n+1)^s}\tag1$$ and $$\displaystyle \sum_{n\geq 1}\frac{1}{n^s}=\frac{1}{2^s}\sum_{n\geq 1}\frac{1}{n^s}+\sum_{n\geq 0}\frac{1}{(2n+1)^s}\tag2$$ then $(1)$ + $(2)$ gives $$\displaystyle \sum_{n\geq 1}\frac{(-1)^n }{n^s}=\left(\frac{1}{2^{s-1}}-1\right)\sum_{n\geq 1}\frac{1}{n^s}=\left(\frac{1}{2^{s-1}}-1\right)\zeta(s)\tag3$$ where $\zeta(\cdot)$ is the Riemann zeta function.

Now differentiating $(3)$ with respect to $s$ gives $$\displaystyle -\sum_{n\geq 1}\frac{(-1)^n \ln n}{n^s}=\left(-\frac{\ln 2}{2^{s-1}}\right)\zeta(s)+\left(\frac{1}{2^{s-1}}-1\right)\zeta'(s)\tag4$$ and letting $s$ tend to $1^+$, gives the result:

$$\displaystyle \sum_{n\geq 1}\frac{(-1)^n }{n}\ln n=-\frac{\ln^2 2}{2}+\gamma \ln 2\tag5$$

where $\gamma$ is the Euler constant and where, near $s=1^+$, we have used $$\frac{1}{2^{s-1}}= 1- (s-1)\ln 2+(s-1)^2\frac{\ln^2 2}{2}+\mathcal{O}((s-1)^3)$$ together with the Laurent series expansions $$ \begin{align} \zeta(s) &=\frac{1}{s-1}+\gamma+\mathcal{O}(s-1)\\\\ \zeta'(s)&=-\frac{1}{(s-1)^2}+\mathcal{O}(1).\end{align}$$

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  • $\begingroup$ The argument in my answer is from an answer to a question that asked why $\gamma$ is the constant term in the Laurent expansion of $\zeta(s)$. We can use that answer to justify the Laurent expansion of $\zeta(s)$ here. $\endgroup$ – robjohn Sep 16 '14 at 23:28
  • $\begingroup$ I believe, however, that the answer should be $\gamma\log(2)-\frac{\log(2)^2}{2}$ $\endgroup$ – robjohn Sep 16 '14 at 23:39
  • $\begingroup$ @robjohn Typo fixed. Thanks! Yes, in this answer, I considered the Laurent expansion for $\zeta(s)$ as a known fact. $\endgroup$ – Olivier Oloa Sep 17 '14 at 5:37

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