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PostPosted: Thu Mar 24, 2016 3:27 pm 

Joined: Fri Dec 04, 2015 4:54 pm
Posts: 14
$$\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$$


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PostPosted: Thu Mar 24, 2016 9:58 pm 

Joined: Sun Dec 13, 2015 2:26 pm
Posts: 56
I am gonna begin this way. Of course, I am sure there are many ways to begin.

Start with the classic:

$$\frac{1}{n^{a}}=\frac{1}{\Gamma(a)}\int_{0}^{\infty}e^{-nt}t^{a-1}dt$$

Then:

$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{a}}=\frac{1}{\Gamma(a)}\sum_{n=1}^{\infty}\cos(nx)\int_{0}^{\infty}e^{-nt}t^{a-1}dt$$

$$=\frac{1}{\Gamma(a)}\int_{0}^{\infty}\left(\sum_{n=1}^{\infty}\cos(nx)e^{-nt}\right)t^{a-1}dt$$

Use the handy and famous Poisson thingie:

$$\sum_{n=1}^{\infty}x^{n}\cos(n\theta)=\frac{x(\cos\theta-x)}{x^{2}-2x\cos\theta+1}$$

and let $x=e^{-t}$ to get:

$$=\frac{1}{\Gamma(a)}\int_{0}^{\infty}\frac{e^{-t}(\cos(x)-e^{-t})}{e^{-2t}-2e^{-t}\cos(x)+1}t^{a-1}dt$$

Make the sub $u=e^{-t}, \;\ t=\ln(1/u), \;\ du=-e^{-t}$

and obtain:

$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^{a}}=-\frac{1}{\Gamma(a)}\int_{0}^{1}\frac{\log^{a-1}(1/u)(u-\cos(x))}{u^{2}-2u\cos(x)+1}du$$

Now diff w.r.t 'a':

$$\sum_{n=1}^{\infty}\frac{\cos(nx)\log(n)}{n^{a}}=\frac{1}{\Gamma(a)}\int_{0}^{1}\frac{(u-\cos(x))\log^{a-1}(1/u)\log(\log(1/u))}{u^{2}-2u\cos(x)+1}du-\frac{\psi(a)}{\Gamma(a)}\int_{0}^{1}\frac{(u-\cos(x))\log^{a-1}(1/u)}{u^{2}-2u\cos(x)+1}du$$

This looks horrendous, but is deceiving. It really ain't too bad upon allowing

$a=1, \;\ x=\pi, \;\ \cos(\pi n)=(-1)^{n}$. Go ahead and make the beginning sum index for the left ln term be n=2 since n=1 results in 0 anyway.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\ln(n)}{n}=\int_{0}^{1}\frac{(u+1)\log\log(1/u)}{u^{2}+2u+1}du-\psi(1)\int_{0}^{1}\frac{u+1}{u^{2}+2u+1}du$$

Since $\psi(1)=-\gamma$, the rightmost integral is elementary and evaluates to $\gamma\ln(2)$

For brevity's sake, let the integral remaining with the loglog be represented by $I$. Resulting in

$$\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n}=I+\gamma \ln(2)$$

Now, the nasty-looking integral with the loglog is not that horrible and is in a class of its own. It has been done quite a bit in the lit and on the forums; all sorts of similar integrals.

This particular one evaluates to $-\frac{1}{2}\ln^{2}(2)$. It is on the sites and in the lit, but can be derived if need be.

So, the final result is then:

$$\boxed{\sum_{n=2}^{\infty}\frac{(-1)^{n}\ln(n)}{n}=-\frac{1}{2}\ln^{2}(2)+\gamma \ln(2)}$$


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PostPosted: Sat Jun 03, 2017 7:48 pm 
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Joined: Mon Oct 26, 2015 12:27 pm
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A solution given by akotronis :

Let $\displaystyle{S_{n}:=\sum_{k=1}^{n}(-1)^k\frac{\ln k}{k}}$. First observe that, by Dirichlet's criterion, the series converges, because $(-1)^k$ has bounded partial sums and $\displaystyle{\frac{\ln k}{k}}$ is eventually decreasing to $0$. Therefore $$\displaystyle\lim_{n\to+\infty}S_{2n}=\sum_{k=1}^{\infty}(-1)^k\frac{\ln k}{k}.$$ Now: \begin{align*}
S_{2n}&=\sum_{k=1}^{n}\frac{\ln2k}{2k}-\sum_{k=1}^{n}\frac{\ln(2k-1)}{2k-1}\\
&=\frac{\ln2}{2}\sum_{k=1}^{n}\frac{1}{k}+\frac{1}{2}\sum_{k=1}^{n}\frac{\ln k}{k}-\left(\sum_{k=1}^{2n}\frac{\ln k}{k}-\sum_{k=1}^{n}\frac{\ln2k}{2k}\right)\\
&=\ln2\,H_{n}+\sum_{k=1}^{n}\frac{\ln k}{k}-\sum_{k=1}^{2n}\frac{\ln k}{k}\\
&=\ln2\,H_{n}-\sum_{k=1}^{n}\frac{\ln(n+k)}{n+k}\\
&=\ln2\,H_{n}-\sum_{k=1}^{n}\frac{\ln n+\ln(1+k/n)}{n+k}\\
&=\ln2\,H_{n}-\ln n\,(H_{2n}-H_{n})-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+k/n}\\
&=H_{n}\ln(2n)-H_{2n}\ln n-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+k/n}\\
&\stackrel{H_{n}=\ln n+\gamma+\mathcal O(1/n)}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\gamma\ln2+\mathcal O(1/n)-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+k/n}\\
&\longrightarrow\gamma\ln2-\int_{0}^{1}\frac{\ln(1+x)}{1+x}\,dx\\
&=\gamma\ln2-\frac{\ln^22}{2}\,.
\end{align*}

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