# Writing sum as integral

I was curious as to what value $\displaystyle\sum_{n\in\mathbb{Z}\text{, }n\ge 2}\frac{(-1)^n}{\log(n)}$ converges to.

One person who commented on this post said that the sum could be written as an integral:

$\displaystyle\int_0^\infty\left(1-\left(1-2^{1-x}\right) \zeta\left(x\right)\right){\rm d}x$

where $\zeta\left(x\right)$ is the Riemann Zeta function.

How do you get from this sum to that integral? And is this indeed the correct integral representation for this particular sum?

• HINT - $\frac{1}{\log(n)}=\int_0^{\infty } n^{-x} \, dx$ – Mariusz Iwaniuk Apr 11 '18 at 17:27
• See this answer math.stackexchange.com/a/476526/61216 – gammatester Apr 11 '18 at 17:35
• I believe it is simpler to write the wanted series just as $$\sum_{n\geq 2}\frac{(-1)^n}{\log n}.$$ – Jack D'Aurizio Apr 11 '18 at 17:55

For any $s\geq 1$ we have $$\zeta(s)=\sum_{n\geq 1}\frac{1}{n^s},\qquad \eta(s)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}=\left(1-\frac{2}{2^s}\right)\zeta(s)$$ but the series defining $\eta(s)$ is conditionally convergent for any $s>0$, allowing the analytic continuation $$\forall s:\text{Re}(s)>0,\qquad \zeta(s)=\left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}=\left(1-\frac{2}{2^s}\right)^{-1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt$$ On the other hand $\frac{1}{\log n}=\int_{0}^{+\infty}\frac{ds}{n^s}$ for any $n\geq 2$, hence $$\sum_{n\geq 2}\frac{(-1)^{n}}{\log n}=\int_{0}^{+\infty}(\eta(s)-1)\,ds=\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{t^{s-1}}{\Gamma(s)e^t(e^t+1)}\,dt\,ds$$ and by integration by parts $$\sum_{n\geq 2}\frac{(-1)^{n}}{\log n}=\iint_{(0,+\infty)^2}\frac{t^s(2+e^{-t})}{\Gamma(s+1)(e^t+1)^2}\,dt\,ds$$ Now the numerical evaluation of the RHS ($\approx 0.9243$) is not difficult since most of the mass of the integral is concentrated near the origin. We may also notice that the "naive" approximation $$\int_{0}^{+\infty}\frac{t^s}{\Gamma(s+1)}\,ds \approx -\frac{1}{2}+\sum_{n\geq 0}\frac{t^n}{n!}=e^t-\frac{1}{2}$$ leads to an accurate estimation: $$\sum_{n\geq 2}\frac{(-1)^n}{\log n}\approx \int_{0}^{+\infty}\frac{(2+e^{-t})\left(-\tfrac{1}{2}+e^t\right)}{(e^t+1)^2}\,dt = \frac{1}{4}+\log 2 = 0.943124\ldots$$ and allows to prove that $\sum_{n\geq 2}\frac{(-1)^n}{\log n}< 1$.