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

Hint : $$ x_n = \frac{\ln{2}}{2} + \frac{\ln{3}}{3} + \cdots \frac{\ln{n}}{n} - \frac{\ln^2{2}}{2} $$ Which converges, if we calculate it's limit we should get $\ln2(\gamma-\frac{\ln{2}}{2})$.

I don't understand where this hint comes from and how it helps us solve the series.

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  • $\begingroup$ How did you solve it? Using $\eta'(1)$? $\endgroup$ – Diger Jan 28 '19 at 15:01
  • $\begingroup$ @Diger See my development of $\eta'(1)$ using only the Euler-Maclaurin Summation Formula. $\endgroup$ – Mark Viola Jan 28 '19 at 16:27
  • $\begingroup$ @SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-) $\endgroup$ – Mark Viola Jan 28 '19 at 16:36
  • $\begingroup$ @Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway. $\endgroup$ – Diger Jan 28 '19 at 17:31
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I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula. To that end we proceed.


Note that we can write any alternating sum $\sum_{n=1}^{2N}(-1)^na_n$ as

$$\sum_{n=1}^{2N}(-1)^na_n=2\sum_{n=1}^N a_{2n}-\sum_{n=1}^{2N}a_n\tag1$$

Using $(1)$, we see that

$$\begin{align} \sum_{n=1}^{2N}(-1)^n \frac{\log(n)}{n}&=2\sum_{n=1}^N \frac{\log(2n)}{2n}-\sum_{n=1}^{2N}\frac{\log(n)}{n}\\\\ &=\log(2)\sum_{n=1}^N\frac1n-\sum_{n=N+1}^{2N}\frac{\log(n)}{n}\tag2 \end{align}$$

Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals

$$\begin{align} \sum_{n=N+1}^{2N}\frac{\log(n)}{n}&=\int_N^{2N}\frac{\log(x)}{x}\,dx+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12 \log^2(2N)-\frac12\log^2(N)+O\left(\frac{\log(N)}{N}\right)\\\\ &=\frac12\log^2(2)+\log(2)\log(N)+O\left(\frac{\log(N)}{N}\right)\tag3 \end{align}$$

Substitution of $(3)$ into $(2)$ yields

$$\begin{align} \sum_{n=1}^{2N}(-1)^n \frac{\log(n)}{n}&=\log(2)\left(-\log(N)+\sum_{n=1}^N \frac1n\right)-\frac12\log^2(2)+O\left(\frac{\log(N)}{N}\right) \end{align}$$

Finally, using the limit definition of the Euler-Mascheroni constant

$$\gamma\equiv\lim_{N\to\infty}\left(-\log(N)+\sum_{n=1}^N\frac1n\right)$$

we arrive at the coveted limit

$$\sum_{n=1}^\infty\frac{(-1)^n\log(n)}{n}=\gamma\log(2)-\frac12\log^2(2)$$

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Here is another method using analytic regularization.

We have $\eta(s)=\left(1-2^{1-s}\right) \zeta(s)$, and so about $s=1$ $$ \eta(s)=\left(\log(2)(s-1) - \frac{\log^2(2)}{2} \, (s-1)^2 + {\cal O}\left((s-1)^3\right)\right) \zeta(s) \\ \zeta(s) = -\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} {\rm d}\lambda \, \lambda^{-s} \, \frac{\rm d}{{\rm d}\lambda} \log \left( \frac{\sin(\pi\lambda)}{\pi \lambda} \right) $$ where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $\lambda=0$, $s<2$ is required also. Substituting $\lambda=it$ $$ \zeta(s) = \frac{\sin\left(\frac{\pi s}{2}\right)}{\pi} \int_{0}^{\infty} {\rm d}t \, t^{-s} \, \frac{\rm d}{{\rm d}t} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) \\ \stackrel{{\rm P.I. | s>1}}{=} \frac{\sin\left(\frac{\pi s}{2}\right)}{\pi (s-1)} \int_{0}^{\infty} {\rm d}t \, t^{1-s} \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) $$ where the second line now converges for $0<s<2$ and hence $$ \eta(s) = \left(\log(2) - \frac{\log^2(2)}{2} \, (s-1) + {\cal O}\left((s-1)^2\right)\right) \frac{\sin\left(\frac{\pi s}{2}\right)}{\pi} \int_{0}^{\infty} {\rm d}t \, t^{1-s} \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) \, . $$ Deriving with respect to $s$ and setting $s=1$ $$ \eta'(1)=-\frac{\log(2)}{\pi} \int_0^\infty {\rm d}t \log(t) \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) - \frac{\log^2(2)}{2\pi} \int_0^\infty {\rm d}t \, \frac{\rm d^2}{{\rm d}t^2} \log \left( \frac{\sinh(\pi t)}{\pi t} \right) \\ =-{\log(2)} \int_0^\infty {\rm d}t \log(t) \, \frac{\rm d}{{\rm d}t} \left( \coth(\pi t) - \frac{1}{\pi t}\right) - \frac{\log^2(2)}{2} $$ we have $$ \coth(\pi t)-\frac{1}{\pi t} = \frac{2t}{\pi} \sum_{k=1}^\infty \frac{1}{k^2+t^2} \, . $$

When interchanging summation and integration order we acquire divergencies, because $\coth(\infty)=1$, but each summand vanishes for $t\rightarrow \infty$. Due to the uniqueness of the result, it does not change up to some divergent part though: $$ -{\log(2)} \int_0^\infty {\rm d}t \log(t) \, \frac{\rm d}{{\rm d}t} \left( \coth(\pi t) - \frac{1}{\pi t}\right) \\ \sim -\log(2) \sum_{k=1}^N \int_0^\infty {\rm d}t \, \log(t) \frac{\rm d}{{\rm d t}} \frac{2t/\pi}{k^2+t^2} \\ =\log(2) \sum_{k=1}^N \int_0^\infty {\rm d}t \, \frac{2/\pi}{k^2+t^2} \\ =\log(2) \sum_{k=1}^N \frac{1}{k} \\ = \log(2) \left\{ \log(N) + \gamma + {\cal O}(1/N) \right\} $$ and therefore $$ \eta'(1)=\gamma \log(2) - \frac{\log^2(2)}{2} \, . $$

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