Here is an interesting topic. It comes from evaluating $$\sum_{k=1}^{\infty}tan^{-1}\left(\frac{1}{k^{2}}\right)$$

I managed to dig up an old paper I have on the sum of arctans by Boros and Moll. It is called the Method of Zeros. It is located here: http://www.mat.utfsm.cl/scientia/archivos/vol11/Art2.pdf on page 6-7

I will post, verbatim, what it says. Perhaps someone can figure out how they arrive at the solution.

They skip over the details and I do not know how they arrived at the general form they establish in [2].

"Based on the factorization of the product $$p_{n}:=\prod_{k=1}^{n}(a_{k}+ib_{k}), \;\ a_{k},b_{k}\in R$$

The argument of $p_{n}$ is given by $$\text{arg}(p_{n})=\sum_{k=1}^{n}\tan^{-1}\left(\frac{b_{k}}{a_{k}}\right)$$

up to an integral multiple of $\pi$.

This can be applied to the case of a polynomial with real coefficients given by


Then, $$\text{arg}(p_{n}(z))=\sum_{k=1}^{n}\tan^{-1}\left(\frac{x-x_{k}}{y-y_{k}}\right)$$

up to an integral multiple of $\pi$.

The special case $$p_{n}(z)=z^{n}-1$$ has roots at

$$z_{k}=\cos(\frac{2\pi k}{n})+i\sin\left(\frac{2\pi k}{n}\right)$$, so we get:

$$\text{arg}(z^{n}-1)=\sum_{k=1}^{n}\tan^{-1}\left(\frac{x-\cos\left(\frac{2\pi k}{n}\right)}{y-\sin\left(\frac{2\pi k}{n}\right)}\right)$$

The classical factorization $$\sin(\pi z)=\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)....[1]$$

shows that one can think of $\sin(\pi z)$ as a polynomial in z of infinite degree.

Euler than derived $$\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}$$ by comparing the cubic terms of [1].

This now yields $$\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{2xy}{k^{2}-x^{2}+y^{2}}\right)=\tan^{-1}(y/x)-\tan^{-1}\left(\frac{\tanh(\pi y)}{\tan(\pi x)}\right)....[2]$$"

That is it. They mention Euler's famous sum of reciprocal of squares then just skip to "this then yields...."

Does anyone see how?. Besides that, didn't Euler compare the square terms and not the cubic ones?.

So, for the problem at hand, we would let $x=y=\frac{1}{\sqrt{2}}$ and arrive at:

$$\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{1}{k^{2}}\right)=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}{\tan\left(\frac{\pi}{\sqrt{2}}\right)}\right)\approx 1.42474......$$

  • $\begingroup$ I think it means the "cubic terms" of the power series in $z$ gotten from $[1]$ to get the sum of the square inverses. Basically, what is the coefficient of $z^3$ on either side of $[1]$? (Note, there is no $z^2$ term in the expansion of the right side of $[1]$. $\endgroup$ – Thomas Andrews Apr 5 '12 at 13:43

I don't understand why you have $x$ in the numerator and $y$ in the denominator in some of the upper equations. However, in $[2]$ they're the right way around. The left-hand side is minus the sum of the arguments of $1-z^2/k^2$, the first term on the right-hand side is the argument of $z$, and the second term on the right-hand side is minus the argument of $\sin(\pi z)$:

$$ \begin{eqnarray} \sin(\pi z) &=& \frac{\mathrm e^{\mathrm i\pi z}-\mathrm e^{-\mathrm i\pi z}}{2\mathrm i} \\ &=& \frac{\mathrm e^{\mathrm i\pi (x+\mathrm iy)}-\mathrm e^{-\mathrm i\pi (x+\mathrm iy)}}{2\mathrm i} \\ &=& \frac{\mathrm e^{-\pi y}(\cos(\pi x)+\mathrm i\sin(\pi x))-\mathrm e^{\pi y}(\cos(\pi x)-\mathrm i\sin(\pi x))}{2\mathrm i} \\ &=& \sin(\pi x)\cosh (\pi y)+\mathrm i\cos(\pi x)\sinh(\pi y)\;. \end{eqnarray} $$

By the way, a lot of your parentheses are undersized.

  • $\begingroup$ Thanks for the response. Nice observation. I do not know why the y's and x's are swapped either. It comes from the paper in the link. I noticed that as well. Perhaps they are typos. Earlier in the paper they bring up $$\sum_{k=1}^{n}tan^{-1}h(k)=tan^{-1}(f(n+1))-tan^{-1}(f(n))$$ where $$h(k)=\frac{f(x+1)-f(x)}{1+f(x+1)f(x)}$$. Perhaps they are using that in some respect. I played around and got close, but did not quite get the $$\frac{2xy}{k^{2}-x^{2}+y^{2}}$$ they obtained. $\endgroup$ – Cody Apr 5 '12 at 15:50
  • $\begingroup$ @Cody: That last expression is the argument of $1-z^2/k^2$, which is the argument of $(k^2-z^2)/k^2$, which is the argument of $k^2-z^2$. $\endgroup$ – joriki Apr 5 '12 at 16:06
  • $\begingroup$ Joriki, if you don't mind, may I ask how you get the arguments of $1-\frac{z^{2}}{k^{2}}$?. I guess I am too stupid to see it. Perhaps I should just forget about it, but I hate it when they skip steps like this. The rest of the paper I understand. $\endgroup$ – Cody Apr 5 '12 at 17:42

Here is a slightly different approach.

Note that $$ \log(1+ix)=\tfrac12\log(1+x^2)+i\tan^{-1}(x)\tag{1} $$ So, applying $(1)$ yields $$ \sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right)=\Im\left(\log\left(\prod_{k=1}^\infty\left(1-\frac{\alpha^2}{k^2}\right)\right)\right)\tag{2} $$ where $\alpha =\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}$, so that $\alpha^2=-i$.

Furthermore, the sinc function has the following product expansion $$ \prod_{k=1}^\infty\left(1-\frac{\alpha^2}{k^2}\right)=\frac{\sin(\pi\alpha)}{\pi \alpha}\tag{3} $$ We also have $$ \begin{align} \frac{\sin(x+iy)}{x+iy} &=\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{x+iy}\\ &=\frac{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}{x^2+y^2}\\ &+\;i\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x^2+y^2}\tag{4} \end{align} $$ Combining $(2)$, $(3)$, $(4)$, and using the identity $\tan(\pi/4-x)=\frac{1-\tan(x)}{1+\tan(x)}$ yields $$ \begin{align} \sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right) &=\Im\left(\log\left(\frac{\sin(\pi\alpha)}{\pi\alpha}\right)\right)\\ &=\tan^{-1}\left(\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}\right)\\ &=\tan^{-1}\left(\frac{\tan(\frac{\pi}{\sqrt{2}})-\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})+\tanh(\frac{\pi}{\sqrt{2}})}\right)\\ &=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right)\tag{5} \end{align} $$

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    $\begingroup$ Very nice. Thanks RobJohn. You always have ingenious solutions. $\endgroup$ – Cody Apr 6 '12 at 10:29
  • $\begingroup$ The underlying assumption is that you remained on the principal branch of the logarithm. $\endgroup$ – Random Variable Apr 6 '12 at 18:38
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    $\begingroup$ @RandomVariable: yes, that is true, but since $|\tan^{-1}(x)|\le|x|$ and $\sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}$, the sum in question is less than $\frac{\pi^2}{6}<\pi$. $\endgroup$ – robjohn Apr 6 '12 at 20:13

I see now. Thanks for the pointers, Joriki.

The roots of $1-\frac{z^{2}}{k^{2}}$ are $-k$ and $k$.

This gives $$\frac{\frac{x-k}{y}+\frac{x+k}{y}}{1+\frac{x-k}{y}\frac{x+k}{y}}$$


  • $\begingroup$ That's a lot more complicated than necessary. You can get the argument directly from the standard formula. The numerator is the imaginary part of $k^2-z^2$, and the denominator is the real part of $k^2-z^2$. $\endgroup$ – joriki Apr 6 '12 at 0:23
  • $\begingroup$ Wow, thanks, Joriki. A case of not seeing the forest for the trees. I should have seen that. blush I was making it too complicated. Thanks a lot for your help and take care. Your help was insightful and valued. $\endgroup$ – Cody Apr 6 '12 at 14:29

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