sum of arctangent 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
$$p_{n}(z)=\prod_{k=1}^{n}(z-z_{k})$$
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:
$$\sum_{k=1}^{\infty}\tan^{-1}\left(\frac{1}{k^{2}}\right)=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}{\tan\left(\frac{\pi}{\sqrt{2}}\right)}\right)\approx 1.42474......$$
 A: 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.
A: 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
$$
\begin{align}
\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{k^2}\right)
&=\Im\left(\log\left(\frac{\sin(\pi\alpha)}{\pi\alpha}\right)\right)\\
&=\tan^{-1}\left(\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}\right)\\
&=\tan^{-1}\left(\frac{\tan(\frac{\pi}{\sqrt{2}})-\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})+\tanh(\frac{\pi}{\sqrt{2}})}\right)\\
&=\frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right)\tag{5}
\end{align}
$$
A: I see now. Thanks for the pointers, Joriki.
The roots of $1-\frac{z^{2}}{k^{2}}$ are $-k$ and $k$.
This gives $$\frac{\frac{x-k}{y}+\frac{x+k}{y}}{1+\frac{x-k}{y}\frac{x+k}{y}}$$
$$=\frac{2xy}{k^{2}-x^{2}+y^{2}}$$.
