Prove that
![\[\sum_{n=1}^{\infty}\arctan \left(\frac{10n}{(3n^2+2)(9n^2-1)}\right) =\ln{3}-\frac{\pi}{4}\]](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-21400b4c687fb75c708e0d8f5764c0df_l3.png "Rendered by QuickLaTeX.com")
Solution
The key ingredient is the observation 
. Then we note that
Using the arg’s property we get that
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*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-3d3feff811e308d21d385fc931046a7b_l3.png "Rendered by QuickLaTeX.com")
since 
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A site of university mathematics
Prove that
Solution
The key ingredient is the observation . Then we note that
Using the arg’s property we get that
Hence the initial sum telescopes;
since which explains why pops up.