Given that:
$\pi = \sum_{k=0}^{\infty} \frac{1}{16^k}\left(\frac{4}{8k+1} - \frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right)$ and $0 \le\left(\frac{4}{8k+1} - \frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right) \le 4$ for all $k$, estimate the number of terms in the series required to approximate $\pi$ within $\frac{1}{1024}$.
I know from the definition of convergence that if we let $s_n =\sum_{k=0}^{n} \frac{1}{16^k}\left(\frac{4}{8k+1} - \frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right)$, that for every $\epsilon >0$ there exists $N \in \mathbb{N}$, such that for all $n\ge N$, $|s_n - \pi|<\epsilon$. We want to find $N$ such that $|s_n-\pi| <\frac{1}{1024}$.
I tried to use the bound to show that $\sum_{k=0}^{N} \frac{1}{16^k}\left(\frac{4}{8k+1} - \frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right) \le \sum_{k=0}^{N} \frac{1}{4^{2k-1}} = \frac{4}{15}\left(1-\frac{1}{16^N}\right)$. But I am not really sure where to go from here since that doesn't really link to the convergence to pi.
Any hints on how to continue?
Hint: What you really want to do is bound $$\sum_{k=N+1}^\infty \left| \frac{1}{16^{k}} \left(\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5}-\frac{1}{8k+6}\right) \right|$$ and find $N$ so that this $< 1/(1024)$.