Originally Posted by

**Showcase_22** Here's what I have so far:

$\displaystyle \pi-4\sum_{k=N}^{\infty}\frac{(-1)^k}{2k+1} \leq 10^{-10}$

$\displaystyle 4 \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}-4 \sum_{k=N}^{\infty} \frac{(-1)^k}{2k+1} \leq 10^{-10}$

$\displaystyle 4\sum_{k=N+1}^{\infty}\frac{(-1)^k}{2k+1} \leq 10^{-10}$

$\displaystyle 4(\frac{-1^{N+1}}{2N+3}+\frac{-1^{N+2}}{2N+5}+\frac{-1^{N+3}}{2N+7}+.........) \leq 10^{-10}$

$\displaystyle 4(\frac{-1^{N+1}}{2N+3}+\frac{-1^{N+2}}{2N+5}+\frac{-1^{N+3}}{2N+7}+.........) \leq 4(\frac{-1^{N+1}}{2N}+\frac{-1^{N+2}}{2N}+\frac{-1^{N+3}}{2N}+......) \leq 10^{-10}$

$\displaystyle 4\sum_{k=N+1}^{\infty}\frac{-1^k}{2k+1} \leq 4\frac{\frac{-1^{N+1}}{2N}}{1-(-1)} \leq 10^{-10}$

$\displaystyle 4\sum_{k=N+1}^{\infty}\frac{-1^k}{2k+1} \leq 4\frac{\frac{-1^{N+1}}{2N}}{2} \leq 10^{-10}$

$\displaystyle 4\sum_{k=N+1}^{\infty}\frac{-1^k}{2k+1} \leq \frac{-1^{N+1}}{N} \leq 10^{-10}$

Solving $\displaystyle \frac{-1^{N+1}}{N} \leq 10^{-10}$ gives:

$\displaystyle -1^{N+1} 10^{10} \leq N$

Therefore $\displaystyle N \geq 10^{10}+1$ so $\displaystyle N=10^{10}+1$ terms.

This seems to be a really big answer. I might just be confusing myself but I was also wondering if there was an easier way of doing it (if this is the right way).

Numerical experiment is always worth a look:

Code:

log_10(N) log_10(|error|)
2 -2.0043
3 -3.0003
4 -4.00004
5 -5.00000
6 -6.00000
7 -7.00000

at which point brute force runs out of umph!

But it looks like a tight bound on the number of terms needed should be ~=10^10

.