Cauchy's Theorem - Prove that $\sum_{n=1}^\infty \frac{1}{\lambda_{n}^2} $ = $\frac{1}{10}$ I seek to prove that $$\sum_{n=1}^\infty \frac{1}{\lambda_{n}^2}   = \frac{1} {10},$$ by applying Cauchy Theorem to $$ f(z) =  \left(\frac{z\tan(z)}{z-\tan(z)}+\frac{3}{z}\right) \frac{1}{z^2},$$ where $\lambda_n$ are positive solutions to $\tan(x) = x$.
How can I solve this? 
 A: Let's see:
$f(z)$ has three types of poles $z_0=0$ , $z_{1,n}=n \pi/2$ and $z_{2,n}=\lambda_n$
It turns out that the poles at $z_{1,n}$ are removeable so the corresponding residue will yield zero (why is that? because $f(z)$ is exactly designed to do so). Now let us integrate $f(z)$  around a contour which consists of two lines just above and below the real axis (connected at infinity), with a sense of orientation which is clockwise. Due to the residue theorem we have 
$$
\oint f(z)dz = -2\pi i [\text{res}(z_0)+\sum_{n=-\infty}^{\infty}\text{res}(z_{2,n})]
$$
But on the other hand, because $tan(z)$ is a bounded function away from the real axis we can also close our contour of integration by adding big semicircles to the now disconnected lines. Because there will no now poles enclosed during this procedure we see that $\oint f(z)dz=0$ and therefore
$$
 -\text{res}(z_0)=\sum_{n=-\infty}^{\infty}'\text{res}(z_{2,n}) \quad (1)
$$
Here $\sum'$ means that we omit $0$ under the sum sign.
The residue at $0$ is given by $\text{res}(z_0)=\frac{1}{5}$, which follows from a straightforward Laurent expansion.
The tricky part know is to calculate the residues at $\lambda_n$. 
First of all because the poles are simple we can use the 
$$
\text{res}(\lambda_n)=\frac{g(\lambda_n)}{f'(\lambda_n)} \quad (2)
$$
with $g(z)=\frac{z^2\tan(z)+3(z- \tan(z))}{z^3}$ and $f'(z)=1-\sec^2(z)=-\tan^2(z)$
using $\tan(\lambda_n)=\lambda_n$ we might rewrite (2)
as
$$
\text{res}(\lambda_n)=-\frac{\lambda_n^2 \lambda_n+ 3(0)}{\lambda_n^3\lambda_n^2}=-\frac{1}{\lambda_n^2}
$$
Putting this into (1) we might observe that
$$
-\frac{1}{5}=-\sum_{n=-\infty}^{\infty}'\frac{1}{\lambda_n^2}
$$
now we observe that the equation $\tan(x)=x$ is anti-symmetric under $x\rightarrow -x$ and all zeros come as pairs $\pm\lambda_n$. This allows us to rewrite
$$
\frac{1}{10}=\sum_{n=1}^{\infty}\frac{1}{\lambda_n^2}
$$
Q.E.D
A: Here is an approach that uses the function $\frac1{\tan(z)-z}$.
First, let's compute the residue of the function $\frac1{\tan(z)-z}$ at $\lambda_n$:
$$
\lim_{z\to\lambda_n}\frac{z-\lambda_n}{\tan(z)-z}=\frac1{\sec^2(\lambda_n)-1}=\frac1{\tan^2(\lambda_n)}=\frac1{\lambda_n^2}\tag{1}
$$
Note that along a contour whose real part is $k\pi$, we have
$$
\tan(k\pi+ix)=i\tanh(x)\tag{2}
$$
whose absolute value is less than $1$. Therefore, along the contour $\gamma$:
$$
k\pi+i[-k,k]\cup\pi[k,-k]+ik\,\cup-k\pi+i[k,-k]\cup\pi[-k,k]-ik\tag{3}
$$
we have $\frac1{\tan(z)-z}=-\frac1z+O\left(\frac1{z^2}\right)$. Thus, the integral along this contour tends to
$$
\int_\gamma\frac{\mathrm{d}z}{\tan(z)-z}=-2\pi i\tag{4}
$$
Near $z=0$,
$$
\begin{align}
\frac1{\tan(z)-z}
&=\frac1{\frac{z^3}3+\frac{2z^5}{15}+O\left(z^7\right)}\\
&=\frac3{z^3}-\frac6{5z}+O(z)\tag{5}
\end{align}
$$
so the residue at $z=0$ is $-\frac65$.
Therefore, combining $(1)$, $(4)$, and $(5)$, we get
$$
\overbrace{\ -2\pi i\ \vphantom{\sum_{n=1}^\infty}}^{\text{integral along $\gamma$}}=\overbrace{-\frac{12\pi i}5\vphantom{\sum_{n=1}^\infty}}^{\text{residue at $z=0$}}+\overbrace{4\pi i\sum_{n=1}^\infty\frac1{\lambda_n^2}}^{\substack{\text{residues at $\lambda_n$}\\\text{in both directions}}}\tag{6}
$$
Solving $(6)$ yields
$$
\sum_{n=1}^\infty\frac1{\lambda_n^2}=\frac1{10}\tag{7}
$$
A: Let's integrate $f$ over a circle of radius $R\gg 1$. The poles of $f$ are at $\pm\lambda_n$ ($f$ is odd, in fact) and $0$. It's easy to see that the residues at $\pm \lambda_n$ equal $-1/\lambda_n^2$, so we obtain $-2\sum 1/\lambda_n^2$ from these. The residue at $z=0$ equals $1/5$; this follows from a calculation, using that
$$
\tan z =z+z^3/3 + 2z^5/15 + O(z^7) .
$$
Note that the first term in parentheses has residue $-3$ at $z=0$; this "explains" the second term.
So
$$
\frac{1}{2\pi i}\int_{C_R} f(z)\, dz = - 2\sum_{\lambda_n<R} \frac{1}{\lambda_n^2} + \frac{1}{5} .
$$
It remains to show that the integral goes to zero as $R\to\infty$. Here we send $R\to\infty$ through values that don't come too close to the $\lambda_n$.
As soon as $\textrm{Im}\, z$ is not too small, $\tan z$ becomes bounded, and thus the integrand is $\lesssim 1/R^2$. Close to the real line, we use what we just agreed on, that $|R-\lambda_n|\gtrsim 1$, where $\lambda_n$ now denotes the nearest $\lambda$. Since $\tan x$ has derivative $1+\lambda_n^2$ at $\lambda_n$, we will have that $|\tan z|\gtrsim \lambda_n^2$ for these $z$, and the integrand becomes $\lesssim 1/R$, which is good enough, since we're using this bound only on a part of bounded length of our circle.
