# Sum of infinite series with arctan

I'm trying to find the value of the following sum (if exist):

$$\sum_{n=1}^{\infty}\left(\arctan\left(\frac{1}{4}-n\right)-\arctan\left(-\frac{1}{4}-n\right)\right)$$ where, $\arctan$ represent the inverse tangent function - $\tan^{-1}$.

I tried to use the telescoping series idea and the sequence of partial sums but I couldn't cancel any terms!

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## 3 Answers

According to Maple, the value is approximately $0.54228694083149264109$. Maple doesn't come up with a closed form, and neither its "identify" function nor the Inverse Symbolic Calculator come up with anything. Do you have any reason to believe there is a closed form for this number?

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No, I have no idea if it has a closed form or not, or even if the sries converge! But if your answer is correct, that's good. If so, is the answer in radian? I mean, is it like $0.54228694083149264109 =0.003012705226841626 \pi$? – Cole Dec 7 '12 at 3:28
Cole, yes it is in radians, and you should not multiply it by $\pi/180$. Your equation is false. It is much closer to $0.1726\pi$, if you want to write it as something times $\pi$. – Jonas Meyer Dec 7 '12 at 3:29
In fact a closed form solution may be found with some work (c.f. my answer). Perhaps that a more direct solution exists... – Raymond Manzoni Dec 8 '12 at 10:38

Consider $$\tag{1}f(x):=\sum_{n=1}^\infty \arctan\left(\left(\frac 14-n\right)x\right)-\arctan\left(\left(-\frac 14-n\right)x\right)$$ and let's rewrite the derivative of $f$ : \begin{align} \tag{2}f'(x)&=\frac 4{x^2}\sum_{n=1}^\infty\frac{1-4n}{(4n-1)^2+\bigl(\frac 4x\bigr)^2}-\frac{-1-4n}{(4n+1)^2+\bigl(\frac 4x\bigr)^2}\\ \tag{3}f'(x)&=\frac 4{x^2}\left(\frac {-1}{1^2+\left(\frac 4x\right)^2}+\sum_{k=1}^\infty\frac{k\sin\bigl(k\frac {\pi}2\bigr)}{k^2+\bigl(\frac 4x\bigr)^2}\right)\\ \end{align} (since the $k=1$ term didn't appear in $(2)$)

But the series in $(3)$ may be obtained from $\,\frac d{d\theta} C_a(\theta)\,$ with : $$\tag{4}C_a(\theta)=\frac {\pi}{2a}\frac{\cosh((\pi-|\theta|)a)}{\sinh(\pi a)}-\frac 1{2a^2}=\sum_{k=1}^\infty\frac{\cos(k\,\theta)}{k^2+a^2}$$ which may be obtained from the $\cos(zx)$ formula here (with substitutions $\ x\to\pi-\theta,\ z\to ia$).

The replacement of the series in $(3)$ by $\,\frac d{d\theta} C_a(\theta)\,$ applied at $\,\theta=\frac {\pi}2$ gives us : \begin{align} f'(x)&=\left(-\arctan\left(\frac x4\right)\right)'-\frac 4{x^2}C_{\frac 4x}\left(\theta\right)'_{\theta=\frac {\pi}2}\\ &=\left(-\arctan\left(\frac x4\right)\right)'+\frac{4\pi}{2x^2}\frac{\sinh\left(\frac{\pi}2 \frac 4x\right)}{\sinh\left(\pi\frac 4x\right)}\\ &=\left(-\arctan\left(\frac x4\right)\right)'+\frac{\pi}{x^2}\frac 1{\cosh\left(\frac{2\pi}x\right)}\\ \end{align}

Integrating both terms returns (with constant of integration $\frac {\pi}2$ since $f(0)=0$) : $$f(x)=\frac {\pi}2-\arctan\left(\frac x4\right)-\arctan\left(\tanh\left(\frac {\pi}x\right)\right)\quad\text{for}\ \ x>0$$ i.e. the neat : $$\tag{5}\boxed{\displaystyle f(x)=\arctan\left(\frac 4x\right)-\arctan\left(\tanh\left(\frac {\pi}x\right)\right)}\quad\text{for}\ \ x>0$$

So that your solution will be (for $x=1$) : $$\boxed{\displaystyle \arctan(4)-\arctan\left(\tanh(\pi)\right)}$$

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Only for the convergence: For every $n \ge 1$ we have \begin{eqnarray} a_n:&=&\arctan\left(\frac14-n\right)-\arctan\left(-\frac14-n\right)\\ &=&\arctan\left(n+\frac14\right)-\arctan\left(n-\frac14\right)\\ &=&\int_{n-\frac14}^{n+\frac14}\frac{1}{1+x^2}dx=\frac{1}{2(1+x_n^2)}, \end{eqnarray} with $$x_n \in \left[n-\frac14,n+\frac14\right] \quad \quad \forall n \ge 1.$$ Therefore $$\tag{1} f(n)\le a_n \le f(-n) \quad \forall n \ge 1,$$ where $$\tag{2} f(z)=\frac{1}{2\left[1+(z+1/4)^2\right]}.$$ It then follows from (1) that the series $\displaystyle\sum_{n=1}^\infty a_n$ converges, and $$S_1:=\sum_{n=1}^\infty f(n)\le \sum_{n=1}^\infty a_n \le S_2:=\sum_{n=1}^\infty f(-n).$$

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