<s>[INST] Compute $\lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{x} {1+k(k+1)x^2}$, $x>0$ I tried using the squeeze theorem, but I had no luck. Then I thought that maybe the sum is telescoping, but also I couldn't see any pattern.
 [/INST] You can use the following formula
\begin{equation}
\arctan{u}+\arctan{v}=\arctan{\frac{u+v}{1-u v}}
\end{equation}
which is valid for $u v <1$. Now consider the sum. For $k=1$ you have
\begin{equation}
\arctan{\frac{x}{1+2 x^2}}
\end{equation}
and for $k=2$
\begin{equation}
\arctan{\frac{x}{1+6 x^2}}
\end{equation}
If you apply the formula above to the first two term you get:
\begin{equation}
\arctan{\frac{x}{1+2 x^2}}+\arctan{\frac{x}{1+6 x^2}}=\arctan{\frac{2x}{1+3 x²}}
\end{equation}
 You can repeat the same computation for the term obtained using $k=3$ and $k=4$ and so on and find the expression:
\begin{equation}
\arctan{\frac{2x}{1+x²(2 m+1)(2m-1)}}
\end{equation}
where the index $m$ runs from 1 to N/2. This procedure can be repeated again and again. For instance the $\arctan{\frac{2x}{1+3 x²}}$ can be added to the sum of the terms obtained by using $k=3$ and $k=4$ and the sum can be written as 
\begin{equation}
\arctan{\frac{4x}{1+x²(4w-31)(4w+1)}}
\end{equation}
where the index $w$ runs from 1 to N/4. By applying the formula of the sum of the two arctan it is possible to compute:
\begin{equation}
\sum_{k=1}^{n} \arctan \frac{x} {1+k(k+1)x^2}=\frac{2 x n}{1+x²(2 n +1)}
\end{equation}
So that
\begin{equation}
\lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{x} {1+k(k+1)x^2}=arccot(x)
\end{equation} 
please note that $arccot(x)=\frac{\pi}{2}-\arctan{x}$.
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