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Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$

How to show $z=\pm\pi$ is a removable singularity for $\frac1{\sin z}+\frac{2z}{z^2-\pi^2}$?

I tried to compute the Laurent series, specifically the coefficients for $1/z,1/z^2,…$ since if we can show those coefficients are all zeroes, we are done. Is there a nice way to compute those coefficients? I tried the contour integral formula but couldn’t get anywhere with it. Thanks.

By the way the latter term can be written as $\frac1{z+\pi}+\frac1{z-\pi}$.

Another related question: how to show $z=0$ is a removable singularity for $\frac1{\sin z}-1/z$?. I tried to write it as $z\left(\frac{z/3!-z^3/5!+z^5/7!-z^7/9!+…}{z-z^3/3!+z^5/5!-…}\right)$ using the Maclaurin series for $\sin z$. But how to show rigorously that the expression can be written as $z\sum_{k=0}^{\infty}a_kz^k$?

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- Indefinite Integral for $\cos x/(1+x^2)$
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- Hardy's inequality again

Related problems: (I), (II). Just take the limit as $z\to \pi$ and $z\to -\pi$ and the two limits should be finite. For instance

$$ \lim_{z\to \pi} \left(\frac1{\sin z}+\frac{2z}{z^2-\pi^2}\right)=\lim_{z\to \pi} \frac{(z^2-\pi^2)+2z\sin(z)}{(z^2-\pi^2)\sin(z)} = \frac{1}{2\pi}. $$

It is just like the function $\frac{\sin(z)}{z}$ which has $z=0$ as a removable singularity, since

$$ \lim_{z\to 0}\frac{\sin(z)}{z}=1. $$

Taking $z = \pi + w$ we get (as $z \to \pi$) $$\eqalign{\frac{1}{\sin(z)} &= \frac{1}{-\sin(w)} = \frac{1}{-w + O(1)} = \frac{-1}{w} + O(1)\cr

\frac{2z}{z^2 – \pi^2} &= \frac{2(\pi + w)}{w^2 + 2 \pi w} = \frac{1+O(w)}{w+O(w^2)} = \frac{1}{w} + O(1)\cr}$$

Thus

$$\frac{1}{\sin(z)} + \frac{2z}{z^2 – \pi^2} = O(1)$$

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