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An arctan series

Prove that

    \[\sum_{n=1}^{\infty}\arctan \left(\frac{10n}{(3n^2+2)(9n^2-1)}\right) =\ln{3}-\frac{\pi}{4}\]


The key ingredient is the observation \arctan x=\arg(1+ix). Then we note that

    \begin{align*} 1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)} &= \frac{\left ( n-i \right )\left ( 3n-\left ( 1-i \right ) \right )\left ( 3n+i \right )\left ( 3n+ \left ( 1+i \right ) \right )}{\left ( 3n-1 \right )\left ( 3n+1 \right )\left ( 3n^2+2 \right )} \\ &= \frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}} \end{align*}

Using the arg’s property we get that

    \begin{align*} \arctan \left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right) &=\arctan\left(\frac1{3n-1}\right)+\arctan \frac{1}{3n}+\\ &\quad  + \arctan\left(\frac1{3n+1}\right)-\arctan \frac{1}{n} \end{align*}

Hence the initial sum telescopes;

    \begin{align*} \sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right) &=\lim_{N \rightarrow +\infty}\sum_{n=1}^N \bigg[ \arctan\left(\frac1{3n-1}\right)+\\ & \quad  + \arctan \frac{1}{3n}+ \arctan \left(\frac{1}{3n+1} \right)- \\ &\quad \quad \quad - \arctan \frac{1}{n} \bigg] \\ &=\lim_{N \rightarrow +\infty}\sum_{n=N+1}^{3N+1}\arctan \frac{1}{n} - \arctan 1\\ &=\lim_{N \rightarrow +\infty} \sum_{n=N+1}^{3N+1}\left[\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right] - \\ &\quad \quad \quad \quad -\arctan 1\\ &=\ln 3-\frac{\pi}{4} \end{align*}

since \displaystyle \sum_{n=N+1}^{3N+1} \frac{1}{n} \sim \ln \left ( \frac{3N+1}{N+1} \right ) which explains why \ln 3 pops up.

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