Approximating $\pi$ with arctangent Use the fact that $\frac{\pi}{4} = \text{arctangent}(\frac{1}{2}) + \text{arctangent}(\frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $\pi$ less than $10^{-3}$.
So far I've got
\begin{align*}
    \pi = 4\Bigg(\sum_{i=1}^{\infty} (-1)^{i+1}\frac{1}{2^{2i-1}(2i-1)} + \sum_{i=1}^{n} (-1)^{i+1}\frac{1}{3^{2i-1}(2i-1)}\Bigg)
\end{align*}
from the given identity. But I don't know how to proceed.

Rewrite the equality as:
\begin{array}{}
    \pi = 4\left[\sum\limits_{i=0}^{n-1} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)+\sum\limits_{i=n}^{\infty} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)\right]
\end{array}
to obtain the inequality to be solved for $n$:
$$
\begin{array}{}
 4\left|\sum\limits_{i=n}^{\infty} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)\right|<\epsilon=10^{-3}.\tag1
\end{array}$$
Observe now that the series (1) is alternating with monotonously decreasing terms, and thus the absolute value of the series is less than that of its first term.
