Let's determine the value of the sum $\sum_{n=1}^{\infty} \operatorname{arcctg}\left(2 n^{2}\right)$.
Solution. Let $\alpha_{n}=\operatorname{arctg}\left(2 n^{2}\right)$ and $\beta_{n}=\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n}$. Then $0<\alpha_{n}<\frac{\pi}{4}$ and $\operatorname{tg} \alpha_{n}=\frac{1}{2 n^{2}}$. Our task is to determine the limit of the sequence $\beta_{n}$.

First, we calculate the value of $\operatorname{tg} \beta_{n}$. We will show by complete induction that $\operatorname{tg} \beta_{n}=1-\frac{1}{n+1}$. If $n=1$, then $\operatorname{tg} \beta_{1}=\operatorname{tg} \alpha_{1}=\frac{1}{2}$. Suppose that $n \geq 2$ and we have already proven the relation $\operatorname{tg} \beta_{n-1}=1-\frac{1}{n}$. Then, based on the addition formula of the tangent function:

$$
\begin{aligned}
\operatorname{tg} \beta_{n} & =\operatorname{tg}\left(\beta_{n-1}+\alpha_{n}\right)=\frac{\operatorname{tg} \beta_{n-1}+\operatorname{tg} \alpha_{n}}{1-\operatorname{tg} \beta_{n-1} \operatorname{tg} \alpha_{n}}=\frac{\frac{n-1}{n}+\frac{1}{2 n^{2}}}{1-\frac{n-1}{n} \cdot \frac{1}{2 n^{2}}}=\frac{\frac{2 n^{2}-2 n+1}{2 n^{2}}}{\frac{2 n^{3}-n+1}{2 n^{3}}}= \\
& =\frac{n\left(2 n^{2}-2 n+1\right)}{(n+1)\left(2 n^{2}-2 n+1\right)}=1-\frac{1}{n+1}
\end{aligned}
$$

Thus, for every positive integer $n$, $\operatorname{tg} \beta_{n}=1-\frac{1}{n+1}$. We also know that $0<\beta_{1}<\frac{\pi}{4}$. If $n \geq 2$ and $0<\beta_{n-1}<\frac{\pi}{4}$, then $0<\beta_{n}=\beta_{n-1}+\alpha_{n}<\frac{\pi}{2}$ and since $\operatorname{tg} \beta_{n}<1$, it follows that $0<\beta_{n}<\frac{\pi}{4}$. Therefore, all elements of the sequence $\beta_{n}$ lie between 0 and $\frac{\pi}{4}$, and $\operatorname{tg} \beta_{n}$ is strictly increasing and approaches 1. Since the tangent function is continuous in the interval $\left(0, \frac{\pi}{2}\right)$, the limit of the sequence $\beta_{n}$ is equal to the arctangent of the limit of the sequence $\operatorname{tg} \beta_{n}$, which is $\operatorname{arctg} 1=\frac{\pi}{4}$. Therefore, the infinite series in the problem is convergent, and its sum is $\frac{\pi}{4}$.