A zeta tail limit

Evaluate the limit

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

Solution

One may use creative telescoping and deduce the double inequality

    \[\frac{1}{n+1}\leq\sum_{k>n}\frac{1}{k^2}\leq \frac{1}{n}\]

since \displaystyle \left(\frac{1}{n}-\frac{1}{n+1}\right)\leq\frac{1}{n^2}\leq \left(\frac{1}{n-1}-\frac{1}{n}\right). The limit follows to be 1.

Another way is using Riemann sum. Note that

    \begin{align*} n \left ( \zeta(2) - \sum_{k=1}^{n} \frac{1}{k^2} \right ) &= n \sum_{k=n+1}^{\infty} \frac{1}{k^2} \\ &= \frac{1}{n} \sum_{k=n+1}^{\infty} \frac{1}{\left ( \frac{k}{n} \right )^2} \\ &\longrightarrow \int_1^{\infty} \frac{{\rm d}x}{x^2}\\ &= 1 \end{align*}

Choose which one you prefer the most.

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