# Closed form $\sum_{n=0}^\infty \frac{\cos(n)}{(2n+1)(2n+2)}$

Is there a closed form for:

$$\sum_{n=0}^\infty \frac{\cos(n)}{(2n+1)(2n+2)}$$

I know it converges through the direct comparison test, and I'm pretty sure there is a closed form, but I'm not entirely certain what it is.

Here's what I've done so far:

$$\sum_{n=0}^\infty \frac{\cos(n)}{2n+1}-\sum_{n=0}^\infty \frac{\cos(n)}{2n+2}$$ $$\sum_{n=0}^\infty \frac{\cos(n)}{2n+1} = \int_{0}^1\sum_{n=0}^\infty x^{2n}\cos(n)dx = \Re \int_{0}^1 \sum_{n=0}^\infty(x^2e^i)^n dx = \Re \int_{0}^1 \frac{1}{1-e^ix^2}dx = \Re({e^{-\frac{i}{2}}\operatorname{arctanh(e^\frac{i}{2}}))}$$

$$\sum_{n=0}^\infty \frac{\cos(n)}{2n+2} = \int_{0}^1\sum_{n=0}^\infty x^{2n+1}\cos(n)dx=\Re\int_{0}^1 x\sum_{n=0}^\infty (x^2e^i)^ndx = \Re \int_{0}^1 \frac{x}{1-e^ix^2}dx=\Re(-\frac{e^{-i}}{2} \ln|1-e^i|)$$

$$\Re({e^{-\frac{i}{2}}\operatorname{arctanh(e^\frac{i}{2}})}+\frac{e^{-i}}{2} \ln|1-e^i|)$$

After plugging all of this in into Wolfram Alpha, I don't get the right answer, however. Where is the mistake? There could be a lot of things wrong.

Edit: I more want to see where my mistake in my work is, rather how to do the problem.

First, we evaluate $\sum_{n=0}^\infty \frac{\cos(n)}{2n+1}$. To do so, we note that for $|z|<1$

$$\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}=\text{arctanh}(z)\tag 1$$

In fact, the series in $(1)$ converges for all $|z|\le 1$, $z\ne 1$.

Therefore, letting $z=e^{ix}$, $x\ne 2k\pi$, reveals

$$\sum_{n=0}^\infty \frac{e^{i(2n+1)x}}{2n+1}=\text {arctanh}(e^{ix})\tag 2$$

for $x\ne 2k\pi$.

Setting $x=1/2$ in $(2)$ yields

$$\text{arctanh}(e^{i/2})=e^{i/2}\sum_{n=0}^\infty \frac{e^{in}}{2n+1}\tag3$$

whence we find

\begin{align} \sum_{n=0}\frac{\cos(n)}{2n+1}&=\text{Re}\left(e^{-i/2}\text{arctanh}(e^{i/2})\right)\\\\ &=\text{Re}\left(e^{-i/2}\frac12\log\left(\frac{1+e^{i/2}}{1-e^{i/2}}\right)\right)\\\\ &=\frac12\text{Re}\left(e^{-i/2}\log\left(i\cot(1/4)\right)\right)\\\\ &=\frac12 \cos(1/2)\log(\cot(1/4))+\frac\pi4 \sin(1/2) \end{align}

Next, we evaluate $\sum_{n=0}^\infty \frac{\cos(n)}{2n+2}=\frac12 \sum_{n=0}^\infty \frac{\cos(n)}{n+1}$. To do so, we note that for $|z|\le 1$, $z\ne1$

\begin{align} \sum_{n=0}^\infty \frac{z^n}{n+1}&=\frac1z \sum_{n=1}^\infty \frac{z^n}{n}\\\\ &=-\frac{\log(1-z)}{z}\tag4 \end{align}

Using $(4)$, we find

\begin{align} \sum_{n=0}^\infty \frac{\cos(n)}{2n+2}&=-\frac12\text{Re}\left(\frac{\log(1-e^{i})}{e^i}\right)\\\\ &=-\frac14 \cos(1)\log(2-2\cos(1))+\frac12\sin(1)\arctan\left(\frac{\sin(1)}{1-\cos(1)}\right)\\\\ &=-\frac12\cos(1)\log(2)-\frac12\cos(1)\log(\sin(1/2))+\sin(1)\left(\frac{\pi-1}{4}\right) \end{align}

For any $z\in\mathbb{C}$ such that $\|z\|\leq 1$ the series $\sum_{n\geq 1}\frac{z^n}{(2n+1)(2n+2)}$ is absolutely convergent.
We may notice that $$f(w)=\sum_{n\geq 0}\frac{w^{2n+2}}{(2n+1)(2n+2)}=\int_{0}^{w}\int_{0}^{v}\sum_{n\geq 1}u^{2n}\,du\,dv=\int_{0}^{w}\int_{0}^{v}\frac{du}{1-u^2}\,dv$$ is a holomorphic function in the region $\|w\|<1$, where it equals $w\,\text{arctanh}(w)+\frac{1}{2}\log(1-w^2)$.
From $$\sum_{n\geq 0}\frac{w^{2n}}{(2n+1)(2n+2)}=\frac{(1+w)\log(1+w)+(1-w)\log(1-w)}{2w^2}$$ we get $$\sum_{n\geq 0}\frac{\rho^{2n} \cos(n)}{(2n+1)(2n+2)}=\text{Re}\frac{(1+\rho e^{i/2})\log(1+\rho e^{i/2})+(1-\rho e^{i/2})\log(1-\rho e^{i/2})}{2\rho^2 e^{i}}$$ for any $\rho\in(0,1)$. By considering the limit of the RHS as $\rho\to 1^-$ and by invoking the dominated convergence theorem we get $$\sum_{n\geq 0}\frac{\cos(n)}{(2n+1)(2n+2)}=\tfrac{1}{2}\cos\tfrac{1}{2}\log\cot\tfrac14+\tfrac12 \cos 1\log\left(2\sin\tfrac12\right)+\tfrac{\pi}{4}\sin\tfrac12+\tfrac{1-\pi}{4}\sin 1.$$ The RHS is approximately $0.513683$.

Hint:

$$\sum_{n=0}^\infty\dfrac{\cos n}{n+2}=$$

Real part of $$e^{-2i}\sum_{n=0}^\infty\dfrac{(e^i)^{n+2}}{n+2}$$

$$=e^{-2i}\left(-\ln(1-e^i)-\dfrac{e^i}{1}\right)$$

and $$\ln(1-e^{2it})=\ln(-i)+it+\ln(2\sin t)$$

• Sorry if this is a stupid question, but how is this helpful to my question? Mar 12, 2018 at 16:18