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What is the exact sum of $$\sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n}$$

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Why do you think there is a simpler way to write this number? –  GEdgar Jan 10 '13 at 20:33
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According to Mathematica, $\gamma\log 2-(\log2)^2/2\approx .1599$. –  Eckhard Jan 10 '13 at 20:33

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up vote 21 down vote accepted

Define $$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^n}{n^s}$$ convergent $\Re(s)>0$, but we shall only use $\eta(s)$ for $\Re(s)>1$. Then the quantity of interest is $$ -\lim_{s \downarrow 1} \frac{\mathrm{d}}{\mathrm{d}s} \eta(s) $$ Now, for $\Re(s)>1$ we can evaluate $\eta(s)$ as follows: $$ \eta(s) + \zeta(s) = \sum_{n=1}^\infty \frac{(-1)^n}{n^s} + \sum_{n=1}^\infty \frac{1}{n^s} = 2 \sum_{m=1}^\infty \frac{1}{(2m)^s} = 2^{1-s} \zeta(s) $$ Hence $$ \eta(s) = \zeta(s) \left(2^{1-s} - 1 \right) $$ Keeping in mind that Riemann function has a pole at $s=1$, $\zeta(s) = \frac{1}{s-1} + \gamma + \mathcal{o}(1)$, where $\gamma$ denotes the Euler-Mascheroni constant we arrive at: $$ \begin{eqnarray} -\lim_{s \downarrow 1} \frac{\mathrm{d}}{\mathrm{d}s} \eta(s) &=& \lim_{s \downarrow 1} \frac{\mathrm{d}}{\mathrm{d}s} \zeta(s) \left(1-2^{1-s} \right) \\ &=& \lim_{s \downarrow 1} \frac{\mathrm{d}}{\mathrm{d}s} \left( \frac{1}{s-1} + \gamma + \mathcal{o}(1) \right) \left(\log(2) (s-1) - \frac{\log^2(2)}{2} (s-1)^2 + \mathcal{o}\left((s-1)^2\right) \right) \\ &=& -\frac{1}{2} \log^2(2) + \log(2) \gamma \end{eqnarray} $$

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(+1) nice answer. –  Mhenni Benghorbal Jan 10 '13 at 21:34
    
similar to a key step in euler's proof of the infinitude of primes –  rondo9 Jan 10 '13 at 22:10

$$\begin{eqnarray}\sum_{k=1}^{2n}(-1)^k\frac{\log(k)}{k} &=& -\sum_{k=1}^{2n}\frac{\log(k)}{k} + \sum_{k=1}^n \frac{\log(2k)}{k}\\ &=& \log(2)\sum_{k=1}^n\frac{1}{k} -\sum_{k=1}^n \frac{\log(n+k)}{n+k}\\ &=& \log(2)\sum_{k=1}^n\frac{1}{k} -\frac{\log(n)}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}- \frac{1}{n}\sum_{k=1}^n\frac{\log(1+\frac{k}{n})}{1+\frac{k}{n}}\\ \end{eqnarray}$$

Now we can recognize two Riemann sums:

$$\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}=\int_1^2\frac{dt}{t} + O(\frac{1}{n}) = \log(2)+O(\frac{1}{n})$$

and

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\log(1+\frac{k}{n})}{1+\frac{k}{n}} =\int_1^2\frac{\log(t)}{t}dt = \frac{1}{2}\log(2)^2.$$

Combining all this gives

$$\begin{eqnarray} \sum_{k=1}^{\infty}(-1)^k\frac{\log(k)}{k}&=&-\frac{1}{2}\log(2)^2+\log(2) \lim_{n\to\infty}\left( \sum_{k=1}^n\frac{1}{k} -\log(n)\right)\\ &=& -\frac{1}{2}\log(2)^2+\log(2)\,\gamma. \end{eqnarray}$$

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Beautiful solution (+1) –  Chris's sis Jan 10 '13 at 22:11

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