I am trying to do the following proof: "show that there exists an integer c such that $$c*\ln(2)- \frac{1}{2} \, (\ln(2))^2 = \sum_{n=1}^{\infty} \frac{ (-1)^n \ln(n)}{n}$$

I have thought of replacing log(n) by the Von Mangolt function, without success. I'm also thinking that I might need to use the Chebychev approximations theorem. i.e.
$$ \frac{c_{1} \, x}{\ln(x)} \leq \pi(x) \leq \frac{c_{2} \, x}{\ln(x)}$$

Thoughts, anyone? J.M.


The key point here is to notice that we can get the factors of $\ln n$ in the numerator by differentiating a Dirichlet series. With

$$\eta(s) = \sum_{n = 1}^{\infty} \frac{(-1)^{n-1}}{n^s}$$

for $\operatorname{Re} s > 0$, we have

$$\eta'(s) = \sum_{n = 1}^{\infty} \frac{(-1)^n\ln n}{n^s}\,,$$

so your series is $\eta'(1)$. Using $\eta(s) = \bigl(1 - 2^{1-s}\bigr)\zeta(s)$, immediately for $\operatorname{Re} s > 1$, then by the identity theorem globally, and the Laurent series of $\zeta(s)$ about $1$,

$$\zeta(s) = \frac{1}{s-1} + \gamma - \gamma_1\cdot (s-1) + \dotsc\,,$$

together with

$$1 - 2^{1-s} = 1 - \exp \bigl(-(s-1)\ln 2\bigr) = (\ln 2)(s-1) - \frac{(\ln 2)^2}{2}(s-1)^2 + \dotsc\,,$$

we obtain the Taylor expansion of $\eta(s)$ about $1$:

\begin{align} \eta(s) &= \biggl((\ln 2)(s-1) - \frac{(\ln 2)^2}{2}(s-1)^2 + \dotsc\biggr) \biggl(\frac{1}{s-1} + \gamma + \dotsc\biggr) \\ &= \biggl(\ln 2 - \frac{(\ln 2)^2}{2}(s-1) + \dotsc\biggr) \bigl(1 + \gamma(s-1) + \dotsc\bigr) \\ &= \ln 2 + \biggl(\gamma\ln 2 - \frac{(\ln 2)^2}{2}\biggr)(s-1) + \dotsc\,. \end{align}

Thus we have

$$\eta'(1) = \gamma \ln 2 - \frac{(\ln 2)^2}{2}$$

and the constant $c$ is not an integer, it is the Euler (or Euler–Mascheroni) constant $\gamma \approx 0.5772$.


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