Example 9 Consider the equation $\frac{(2 k+1)-(2 k-1)}{1+(2 k+1)(2 k-1)}=\frac{1}{2 k^{2}}$ and its resemblance to the trigonometric formula $\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}$.
We can write it in the form of the arctangent function: $\arctan \frac{1}{2 k^{2}}=\arctan (2 k+1)-\arctan (2 k-1), k \in$ $\mathbf{N}^{*}$.
Thus, we have the identity:
$\arctan \frac{1}{2}+\arctan \frac{1}{8}+\arctan \frac{1}{18}+\cdots+\arctan \frac{1}{2 \cdot n^{2}}=\arctan (2 n+1)-\arctan 1$.
Similarly, we consider the identity:
$$
\frac{\frac{k}{k+1}-\frac{k-1}{k}}{1+\frac{k}{k+1} \cdot \frac{k-1}{k}}=\frac{1}{2 k^{2}} \text {. }
$$

Writing it in the form of the arctangent function:
$\arctan \frac{1}{2 k^{2}}=\arctan \frac{k}{k+1}-\arctan \frac{k-1}{k}, k \in \mathbf{N}^{*}$.
We can also obtain the identity:
$\arctan \frac{1}{2}+\arctan \frac{1}{8}+\arctan \frac{1}{18}+\cdots+\arctan \frac{1}{2 \cdot n^{2}}=\arctan \frac{n}{n+1}$.