I would like to approximate the infinite sum $$\sum_{k=1}^{\infty}\frac{1}{k^2}$$ up to a precision of $10^{-6}$. For that I want to know at what $m$ I can stop, so that the remaining part will be insignificant. So I would like to find an $m$ such that $$\sum_{m+1}^{\infty}\frac{1}{k^2}<10^{-6}$$
I would welcome any hints or links.
Hard to do a lot better than the integral.
As $f(x)=\frac 1{x^2}$ is monotonically decreasing, we get $$\int_m^{\infty}\frac {dx}{x^2}>\sum_{m+1}^{\infty}\frac 1{k^2}>\int_{m+1}^{\infty}\frac {dx}{x^2}$$
Thus $$\frac 1{m}>\sum_{m+1}^{\infty}\frac 1{k^2}>\frac 1{m+1}$$
Taking $m=10^6$ shows that $$\frac 1{10^6}>\sum_{10^6+1}^{\infty}\frac 1{k^2}>\frac 1{10^6+1}$$ So that value suffices but no smaller value will.
Worth remarking that in this case, the exact value of the infinite sum is known (it is $\frac {\pi^2}6$) so there are better ways to approximate it.
