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.
On
The sums are alternating and the terms decreasing in absolute value. This implies that the error is lesser or equal to the first discarded term. The error summing up to $n$ is then lesser or equal to $$ 4\Bigl(\frac{1}{2^{2n+1}(2\,n+1)}+\frac{1}{3^{2n+1}(2\,n+1)}\Bigr). $$ Now choose $n$ so that the above is $<10^{-3}$.
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.