Telescoping with arctangents and products

1. $1-\frac1{k^2}=\frac{k-1}k\cdot\frac{k+1}k$, so
$$\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2=\frac{n+1}{2n}.$$

2. Key identity. Let $\alpha=\arctan\frac1{2k-1}$ and $\beta=\arctan\frac1{2k+1}$, both in $\left(0,\frac\pi2\right)$ with $\alpha>\beta$. Then
$$\tan(\alpha-\beta)=\frac{\frac1{2k-1}-\frac1{2k+1}}{1+\frac1{(2k-1)(2k+1)}}=\frac1{2k^2}.$$
Since $\alpha-\beta\in\left(0,\frac\pi2\right)$, we get $\alpha-\beta=\arctan\frac1{2k^2}$.

Summing.
$$\sum_{k=1}^n\arctan\frac1{2k^2}=\arctan1-\arctan\frac1{2n+1}=\frac\pi4-\arctan\frac1{2n+1}.$$
As $n\to\infty$, the sum tends to $\dfrac\pi4$.

Check. $n=1$: $\arctan\frac12=\frac\pi4-\arctan\frac13$ ✓.

Lesson. Telescoping: try to write each term as $f(k)-f(k+1)$, using factorisations (products) or addition formulas (arctangents).