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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.

Tolaso
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2 Answers2

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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.

Community
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robjohn
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  • and the term $O\left(\frac{\ln(n)}{n}\right)$ comes from $\frac{\ln(n)}{2n}$. Is it correct? Thank you in advance! – Ricardo770 Oct 15 '21 at 19:51
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    @Ricardo770: Euler-Maclaurin says that $$\sum_{k=1}^n\frac{\log(k)}k=\tfrac12\log(n)^2+\frac{\log(n)}{2n}+\gamma_1+\int_n^\infty\left({x}-\tfrac12\right)\frac{\log(x)-1}{x^2},\mathrm{d}x$$ where $\gamma_1=-0.0728158454836767$ is a Stieltjes constant and it is not hard to show that for $n\ge5$, $$-\frac{\log(n)-1}{4n^2}\le\int_n^\infty\left({x}-\tfrac12\right)\frac{\log(x)-1}{x^2},\mathrm{d}x\le0$$ and $$\gamma_1=\int_1^\infty\left({x}-\tfrac12\right)\frac{1-\log(x)}{x^2},\mathrm{d}x$$ – robjohn Oct 16 '21 at 17:17
  • thank you very much for you reply. I am a bit confused, shouldn´t the last term in the first order E-M be $\int_1^\infty f^\prime(x) P_1(x),dx-\int_n^\infty f^\prime(x) P_1(x),dx$ , and in this case $f(x)=\frac{\ln(x)}{x}$. Thank you again. – Ricardo770 Oct 17 '21 at 10:02
  • @Ricardo770: look at the equation for $\gamma_1$ and the first equation in my previous comment. – robjohn Oct 17 '21 at 10:32
  • thank you very much. – Ricardo770 Oct 17 '21 at 13:37
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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 $$ \begin{align} \zeta(s) &=\frac{1}{s-1}+\gamma+\mathcal{O}(s-1)\\\\ \zeta'(s)&=-\frac{1}{(s-1)^2}+\mathcal{O}(1).\end{align}$$

Olivier Oloa
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  • 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. – robjohn Sep 16 '14 at 23:28
  • I believe, however, that the answer should be $\gamma\log(2)-\frac{\log(2)^2}{2}$ – robjohn Sep 16 '14 at 23:39
  • @robjohn Typo fixed. Thanks! Yes, in this answer, I considered the Laurent expansion for $\zeta(s)$ as a known fact. – Olivier Oloa Sep 17 '14 at 05:37