Problem source

Prove that  
$\displaystyle \arctan a + \arctan b = \arctan \l {a + b \over 1-ab} \r\,$
when $0 < a < 1$ and $0 < b < 1\,$.
Prove by induction that, for $n \ge 1\,$,
\[
\sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r
\]
and hence find
\[
\sum_{r = 1}^\infty \arctan \l {1 \over r^2 + r + 1} \r\,.
\]
Hence prove that 
\[
\sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r = {\pi \over 2}\,.
\]

Solution source

\begin{align*}
&& \arctan a &\in (0, \tfrac{\pi}{4}) \\
&& \arctan b &\in (0, \tfrac{\pi}{4}) \\
\Rightarrow && \arctan a+\arctan b &\in (0, \tfrac{\pi}{2}) \\
&& \tan \left ( \arctan a+\arctan b \right) &= \frac{\tan \arctan a + \tan \arctan b}{1 - \tan \arctan a \tan \arctan b} \\ 
&&&= \frac{a+b}{1-ab} \in (0, \infty) \\
\Rightarrow && \arctan \left ( \frac{a+b}{1-ab} \right) &\in (0, \tfrac{\pi}{2}) \\
\Rightarrow && \arctan a + \arctan b &= \arctan \left ( \frac{a+b}{1-ab} \right)
\end{align*}

Claim: $\displaystyle \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r$

Proof: (By Induction):

Base case ($n=1$): 
\begin{align*}
&& LHS &= \sum_{r=1}^1 \arctan \left ( \frac{1}{r^2+r+1} \right) \\
&&&= \arctan \left ( \frac{1}{3} \right) \\
&& RHS &= \arctan \left ( \frac{1}{1+2} \right)\\
&&&= \arctan \left ( \frac{1}{3} \right) = LHS
\end{align*}

Inductive step, suppose true for $n = k$, ie

\begin{align*}
&& \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r &= \arctan \l {k \over k+2} \r \\
\Rightarrow && \sum_{r = 1}^{k+1} \arctan \l {1 \over r^2 + r + 1} \r &= \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r+ \arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\
&&&= \arctan \l {k \over k+2} \r+\arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right)  \\
&&&= \arctan \left ( \frac{{k \over k+2}+\frac{1}{(k+1)^2+(k+1)+1} }{1-\frac{k}{k+2}\frac{1}{(k+1)^2+(k+1)+1} } \right) \\
&&&= \arctan \left ( \frac{k((k+1)^2+k+1+k)+(k+2) }{(k+2)((k+1)^2+(k+1)+1)-k} \right) \\
&&&= \arctan \left ( \frac{k^3+3k^2+4k+2 }{k^3+5k^2+8k+6} \right) \\
&&&= \arctan \left ( \frac{(k+1)(k^2+2k+2) }{(k+3)(k^2+2k+2)} \right) \\
&&&= \arctan \left ( \frac{k+1 }{(k+1)+2} \right) \\
\end{align*}

Therefore it is true for $n = k+1$, therefore it is true for all $n \geq 1$ by the principle of mathematical induction.

\begin{align*}
&& S &= \lim_{n \to \infty} \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r \\
&&&= \lim_{n \to \infty} \arctan \l \frac{n}{n+2} \r \\
&&&= \lim_{n \to \infty} \arctan \l \frac{1}{1+2/n} \r \\
&&&=\arctan\l \lim_{n \to \infty}   \frac{1}{1+2/n} \r \\
&&&= \frac{\pi}{4}
\end{align*}

\begin{align*}
&& \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r &= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ (r+1)^2 - (r+1) + 1} \right) \\
&&&= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ r^2+r+1} \right) \\
&&&= \arctan \l \frac{1}{0^2+0+1} \r + \frac{\pi}{4} \\
&&&= \frac{\pi}{2}
\end{align*}