# Calculate the difference between π/4 and the Leibniz series for computing &#x03C0;<!-- π -->

Calculate the difference between π/4 and the Leibniz series for computing $\pi /4$ with $n=200$.
This series appears to converge relatively slowly, and so at what point can we confidently say that "the $576$th digit is $3$" of an irrational number?
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postojahob
The way you can tell how many digits you have computed is by providing a bound on the remainder term for the series (i.e., of ${R}_{k}=\sum _{n=k+1}^{\mathrm{\infty }}{a}_{n}$). Say you want m digits, then you want ${10}^{-m}>|{R}_{k}|$.
For the Liebniz series $\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+1}$, since it is an alternating series, we know that the remainder term is bounded by $\frac{1}{2n+1}$ itself, and so if you want $m$ digits of $\frac{\pi }{4}$, $\frac{{10}^{m}-1}{2}$ terms would be sufficient (quite a few).
For other series, sometimes Tayor's theorem can be used to provide a bound on the remainder term.