Remember your inverse formulae!

Let $\theta_n = \displaystyle \sum_{r=1}^n \arctan \left(\frac 1{2r^2} \right)$ . Then we have:

$\begin{aligned} \theta_1 & = \arctan \left(\frac 12\right) \\ \theta_2 & = \arctan \left(\frac 12\right) + \arctan \left(\frac 18\right) = \arctan \left(\frac {\frac 12 + \frac 18}{1-\frac 12 \cdot \frac 18} \right) = \arctan \left(\frac 23 \right) \\ \theta_3 & = \arctan \left(\frac 23 \right) + \arctan \left(\frac 1{18} \right) = \arctan \left(\frac {\frac 23 + \frac 1{18}}{1-\frac 23 \cdot \frac 1{18}} \right) = \arctan \left(\frac 34 \right) \end{aligned}$

It appears that $\theta_n = \arctan \left(\dfrac n{n+1} \right)$ and we can prove it by induction that the claim is true for all $n \ge 1$ . We know that the claim is true for $n=1$ . Assume that it is true for $n$ , then:

$\begin{aligned} \theta_{n+1} & = \theta_n + \arctan \left(\frac 1{2(n+1)^2} \right) \\ & = \arctan \left(\frac n{n+1} \right) + \arctan \left(\frac 1{2(n+1)^2} \right) \\ & = \arctan \left(\frac {\frac n{n+1} + \frac 1{2(n+1)^2}} {1- \frac n{n+1} \cdot \frac 1{2(n+1)^2}} \right) \\ & = \arctan \left(\frac {(n+1)(2n^2+2n+1)}{2(n+1)^3 - n} \right) \\ & = \arctan \left(\frac {(n+1)(2n^2+2n+1)}{2(n^3+3n^2+3n+1) - n} \right) \\ & = \arctan \left(\frac {(n+1)(2n^2+2n+1)}{2n^3+6n^2+5n+2} \right) \\ & = \arctan \left(\frac {(n+1)(2n^2+2n+1)}{(n+2)(2n^2+2n+1)} \right) \\ & = \arctan \left(\frac {n+1}{n+2} \right) \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ .

Then we have $\displaystyle \lim_{n \to \infty} \sum_{r=1}^n \arctan \left(\frac 1{2r^2} \right) = \lim_{n \to \infty} \arctan \left(\frac n{n+1} \right) = \arctan (1) = \frac \pi 4 = \boxed{45}^\circ$ .

$\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \arctan { \left( \frac { 1 }{ 2{ r }^{ 2 } } \right) } } } =\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \arctan { \left( \frac { 2 }{ 4{ r }^{ 2 } } \right) } } } =\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \arctan { \left( \frac { (2r+1)-(2r-1) }{ 1+(2r+1)(2r-1) } \right) } } } =\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \arctan { (2r+1) } -\arctan { (2r-1) } } } =\lim _{ n\rightarrow \infty }{ \arctan { (2n+1) } -\arctan { 1 } } =90-45=45$

this link tells us that it sums to pi/4
this link tells us that pi/4 equals 45 degrees