An infinite sum of Inverse Tangents

$\DeclareMathOperator{\sech}{sech}$ 

I want to discuss the following sum 

$$\sum_{i=0}^\infty (-1)^{n}\tan^{-1}(\frac{x}{2n+1})$$

Using a few neat tricks we will be able to simplify this into a closed form expression. First we can quickly check that the sum exists. Note that this is an alternating series, and that the term $$\lim_{n\rightarrow \infty} \tan^{-1}(\frac{x}{2n+1}) \rightarrow 0$$

To evaluate this sum, lets begin by examining the first term, $\tan^{-1}(x)$. Cotangent can be written in terms of logarithms.

$$\tan^{-1}(x) = \frac{i}{2}( \ln(x+i) - \ln(x-i) )$$ Notice that this has logarithmic singularities at $x = \pm i$. Also notice that the distance between these two singularities is $2$. If we examine the next term in the sequence we have $$-\tan^{-1}(\frac{x}{3}) = \frac{i}{2}(-\ln(\frac{x}{3}+i) + \ln(\frac{x}{3}-i) )$$ Notice that this has singularities with opposite polarity at $x=\pm 3i$. Moreover, the distance between these singularities and the corresponding singularities of $\tan^{-1}(x)$ are also 2. Thus we can write our some as an alternating sums of logarithms with singularity at the odd imaginary integers. That is

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

Note, we can multiply or divide the top the inside of the logarithm as 

$$\ln(a(x+i)) - \ln(a(x-i))= \ln(a) +\ln(x+i) - \ln(a)- \ln(x-i)= \ln(x+i)-\ln(x-i)$$

If we take the derivative (remembering to integrate afterwards) we can see this series is a hyperbolic secant in disguise. 

Note that

  $$\frac{d}{dx}\frac{i}{2} \sum_{i=-\infty}^\infty (-1)^{n}\ln(x - (2n +1)i) =  \frac{i}{2}\sum_{i=-\infty}^\infty (-1)^{n}\frac{1}{x-(2n+1)i}=$$

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

This sum should ring a bell. It is the partial fraction sum for hyperbolic secant. We have discussed it before here, and can be found on Wikipedia here and a proof here. 

To summarize $$\frac{1}{2}\sum_{i=-\infty}^\infty (-1)^{n}\frac{1}{i x-(2n+1)}= \frac{\pi}{4}\sech(\frac{\pi x}{2})$$ Now, our series has been added up. However, now we need to undo our derivative, and integrate. We also will note when $x=0$ our sum equals $0$, and we will use this constraint to determine the integration constant $C$. Note that

$$\int \sech(x) dx = 2 \tan^{-1} (\tanh(\frac{x}{2}) ) +C$$ , therefore

$$\int \frac{\pi }{4}\sech(\frac{\pi x}{2})dx =  \tan^{-1}(\tanh(\frac{\pi x}{4}))+C$$ Because the expression needs to equal 0 at $x=0$ $C=0$. This gives us our final expression of:

$$\sum_{i=0}^\infty (-1)^{n}\tan^{-1}(\frac{x}{2n+1}) = \tan^{-1}(\tanh(\frac{\pi x}{4}))$$

Author: Alexander White