# Equality with Euler–Mascheroni constant

While trying to prove integral with exponential function and logarithm in an alternative way, I came to this solution:

$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}.$$

As both solutions have to be the same, the following equality should be valid:

$$\sum_{k=0}^{+\infty}(-1)^{k+1}\frac{\log (k+1)+\gamma }{(k+1)}=- \frac{1}{2}{{\log }^2(2)}.$$

Can anyone give me some advice on how to prove this equality.

p.s. You can be sure that the equality is correct, as I checked it numerically.

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So, obviously the hard part is $\displaystyle \sum_k \frac{(-1)^k\log(k)}{k!}$. I think a possible route would be to see if you can find $\displaystyle \sum_k \frac{(-1)^kk^x}{k!}$ for a general $x$ (it's easy for integral $x$) and then you can differentiate both sides and let $x=0$. I have to run, but that should work, assuming that sum isn't actually too crazy for non-integral $x$. –  Alex Youcis Jan 20 '12 at 23:26
Are you sure you wrote this right? I get $\sum_{k=0}^\infty (-1)^{k+1} \frac{\log(k+1)+\gamma}{(k+1)!} = -.1548995048$ and $-\frac{1}{2} \log^2(2) = -.2402265070$ approximately –  Robert Israel Jan 20 '12 at 23:39
The ! should be removed from the denominator: $\sum_{k=0}^{\infty} (-1)^{k+1} \frac{\log(k+1)+\gamma}{k+1}=-\frac{1}{2}(\log 2)^2.$ This is obtained from the integral by expanding $(1+e^x)^{-1}=e^{-x}-e^{-2x}+{\dots}$ and integrating term-by-term. –  David Moews Jan 20 '12 at 23:52
This is really annoying. I removed the !. I am very ver sorry for this error. –  wnvl Jan 21 '12 at 0:18

Note that \begin{align} \sum_{k=1}^\infty(-1)^k\frac{\log(k)+\gamma}{k} &=\lim_{n\to\infty}\left(2\sum_{k=1}^{n}\frac{\log(2k)+\gamma}{2k}-\sum_{k=1}^{2n}\frac{\log(k)+\gamma}{k}\right)\\ &=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{\log(2)+\log(k)+\gamma}{k}-\sum_{k=1}^{2n}\frac{\log(k)+\gamma}{k}\right)\tag{1} \end{align} Using the Euler-Maclaurin Sum Formula, we get that $$\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+O\left(\frac{1}{n^4}\right)\tag{2}$$ and that $$\sum_{k=1}^n\frac{\log(k)}{k}=\frac12\log(n)^2+C+\frac{\log(n)}{2n}-\frac{\log(n)-1}{12n^2}+O\left(\frac{\log(n)}{n^4}\right)\tag{3}$$ Applying $(2)$ and $(3)$ to $(1)$, leaving out the terms which vanish, we get \begin{align} &\sum_{k=1}^\infty(-1)^k\frac{\log(k)+\gamma}{k}\\ &=\small{\lim_{n\to\infty}\left(\log(2)(\log(n){+}\gamma)+\left(\frac12\log(n)^2+C+\gamma(\log(n){+}\gamma)\right)-\left(\frac12\log(2n)^2+C+\gamma(\log(2n){+}\gamma)\right)\right)}\\ &=\lim_{n\to\infty}\left(\log(2)(\log(n)+\gamma)-\log(2)\log(n)-\frac12\log(2)^2-\gamma\log(2)\right)\\ &=-\frac12\log(2)^2\tag{4} \end{align}
@wnvl: The error in $(3)$ turns out to be less than $\dfrac{log(n)}{120n^4}$, so we can compute $C = -0.07281584548367672486$ to $20$ places by setting $n=100000$. –  robjohn Jan 21 '12 at 17:22
This can be solved similarly to the original problem. The Dirichlet eta function is defined by $$\eta(s):=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n^s}.$$ The given sum can be rewritten as $$\sum_{n\ge 1} (-1)^n \frac{\log n}{n}+\sum_{n\ge 1} (-1)^n \frac{\gamma}{n}= \eta'(1)-\gamma \log 2.\qquad (*)$$ We have $$\eta(s)=\sum_{n\ge 1} \frac{1}{n^s}-2\sum_{n\ge 1} \frac{1}{(2n)^s}=(1-2^{1-s})\zeta(s)$$ so, using the expansions $$\zeta(s)=\frac{1}{s-1}+\gamma+O(s-1),$$ $$2^{1-s}=e^{(1-s)\log 2}=1-(s-1)\log 2+\frac{1}{2}(\log 2)^2 (s-1)^2+O((s-1)^3),$$ we get $\eta'(1)=\gamma \log 2 -\frac{1}{2}(\log 2)^2$, so (*) equals $-\frac{1}{2}(\log 2)^2$.
In fact, I use the result from my answer to derive the expansion for $\zeta(s)$ you cite above. See Riemann zeta near 1 –  robjohn Jan 28 '12 at 14:16