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ζ(3) tail in floor function

Let n \in \mathbb{N}. Prove that

    \[\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1)\]

Solution

We note that

    \begin{align*} \sum_{k=n}^{\infty} \frac{1}{k^3} &< \sum_{k=n}^{\infty} \frac{1}{k(k^2-1)} \\ &= \sum_{k=n}^{\infty} \left \frac{1}{2k(k-1)} - \frac{1}{2k(k+1)}\right) \\ &= \frac{1}{2n(n-1)} \end{align*}

as well as

    \begin{align*} \sum_{k=n}^{\infty} \frac{1}{k^3} &> \sum_{k=n}^{\infty} \frac{k}{k^4 + \frac{1}{4}} \\ &= \sum_{k=n}^{\infty} \left( \frac{1}{2k^2 -2k + 1} - \frac{1}{2(k+1)^2 - 2(k+1) + 1}\right) \\ & =\frac{1}{2n^2 - 2n + 1} \\ &= \frac{1}{2n(n-1) + 1} \end{align*}

The result follows.

Note: It also holds that

    \[\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^2}\right)^{-1}\right\rfloor=n-1\]

Using the same exact technique we see that

    \[\sum_{k=n}^{\infty}\frac {1}{k^2}> \sum_{k=n}^{\infty}\frac{1}{k(k+1)}=\sum_{k=n}^{\infty}\left(\frac 1k-\frac{1}{k+1}\right)=\frac{1}{n}\]

as well as

    \[\sum_{k=n}^{\infty}\frac {1}{k^2}< \sum_{k=n}^{\infty}\frac{1}{k(k-1)}=\sum_{k=n}^{\infty}\left(\frac 1{k-1}-\frac{1}{k}\right)=\frac {1}{n-1}\]

The result follows.

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