Approximating the sum of a series to a given error

Question

Find the minimum $n$ such that $\displaystyle\sum_{k=n+1}^\infty \dfrac{1}{k^3}<10^{-6}.$

I know there's no exact formula for $\zeta(3)$ and I tried approximating the sum by using a Taylor series, but I can't seem to find a way to show that $\displaystyle\sum_{k=708}^\infty \dfrac{1}{k^3}<10^{-6}.$ However, I do know how to show that $\displaystyle\sum_{k=707}^\infty \dfrac{1}{k^3}>10^{-6}$ using an integral approximation.

Answer 1

Show that for $ k \geq 1, \frac{1}{k^3 } < \frac{1}{2(k-\frac{1}{2} ) ^2 } - \frac{1}{2(k+ \frac{1}{2} )^2} $.

Hence, from telescoping series,

$$ \sum_{k=707+1}^\infty \frac{1}{k^3} < \sum_{k=707+1}^\infty \frac{1}{2(k-\frac{1}{2} ) ^2 } - \frac{1}{2(k+ \frac{1}{2} )^2} = \frac{1}{2 (707.5)^2} < \frac{1}{10^6}. $$

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Note: This could be expressed as an integral inequality because $f(x) = \frac{1}{k^3 } $ is convex, so the midpoint approximation is an under estimate. In other words:

$$ \frac{1}{k^3} < \int_{k-\frac{1}{2} } ^ {k+\frac{1}{2} } \frac{1}{k^3} \, dk = [\frac{-1}{2k^2}]_{k-\frac{1}{2}} ^ {k+\frac{1}{2} } = \frac{1}{2(k-\frac{1}{2} ) ^2 } - \frac{1}{2(k+ \frac{1}{2} )^2} .$$

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The case for $ n = 706$ follows directly from

$$ \frac{1}{2n^2} = \int_{n}^{\infty} \frac{1}{k^3} \, dk < \sum_{k=n}^\infty \frac{1}{k^3}$$

Answer 2

We have that $\int_{n+1}^\infty \frac1{x^3}dx< \sum_{k=n+1}^\infty \frac1{k^3}< \int_n^\infty \frac1{x^3}dx,$ so calculating the integrals, we have $\frac1{2(n+1)^2}<\sum_{k=n+1}^\infty \frac1{k^3}<\frac1{2n^2}.$ So if $n$ is minimal such that $\frac1{2(n+1)^2}<\sum_{k=n+1}^\infty \frac1{k^3}<10^{-6},$ we have that $2(n+1)^2>10^6,$ or equivalently, $n>\sqrt{10^6/2}-1>706.$

Since $n$ is minimal, we also have that $\frac1{2(n-1)^2}>\sum_{k=(n-1)+1}^\infty \frac1{k^3}>10^{-6},$ or $n<\sqrt{10^6/2}+1<709.$

Hence, we must have $n=707$ or $708.$

All that is left to do is show $\sum_{k=708}^\infty \frac1{k^3}<10^{-6},$ but we have $\sum_{k=708}^\infty \frac1{k^3}=\sum_{k=708}^{766} \frac1{k^3}+\sum_{k=767}^\infty \frac1{k^3}<\sum_{k=708}^{766} \frac1{k^3}+\frac1{2\cdot 766^2}<0.99999996944\cdot10^{-6}<10^{-6}.$

Answer 3

If you know the polylogarithm function $$S_n=\sum_{k=n+1}^\infty \dfrac{1}{k^3}=-\frac{\psi ^{(2)}(n+1)}{2}$$ Using their asymptotics $$S_n=\frac{1}{2 n^2}-\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$ So, a first approximation is $$\frac{1}{2 n^2}=10^{-6} \implies n=500 \sqrt{2}\implies n > 707\implies n=708$$