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An arctan series

Evaluate the series

    \[\Omega= \sum_{n=1}^{\infty} \arctan \left ( \frac{9}{9+(3n+5)(3n+8)} \right )\]

(Dan Sitaru)

Solution

Well,

    \begin{align*} \Omega &= \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\arctan\left(\frac{9}{9+(3k+5)(3k+8)}\right)\\ &= \lim_{n\rightarrow\infty}\sum_{k=1}^{n}\left[\arctan\left(k+\frac{8}{3}\right)-\arctan\left(k+\frac{5}{3}\right)\right]\\ &= \lim_{n\rightarrow\infty}\left[\arctan\left(n+\frac{8}{3}\right)-\arctan\left(\frac{8}{3}\right)\right]\\ &=\frac{\pi}{2}-\arctan\frac{8}{3}=\arctan\frac{3}{8}. \end{align*}

and the problem is over.

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