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For each $k$, $$\sum _{n=k^2+1}^{k^2+2 k} \left(\sqrt{n}-k\right)\approx \frac{6 k+1}{6}$$

A generating function $f(k)$ that removes $k$ from the output, retaining only the fractional part, $$f(k)\equiv \zeta \left(-\frac{1}{2},k^2+1\right)-\zeta \left(-\frac{1}{2},(k+1)^2\right)-k (2 k+1)$$
where $\zeta \left(a,b\right)$ is the Hurwitz Zeta function.

The limit increases precision very slowly, $$\lim_{k\to \infty } \, f(k)\approx \frac{1}{6}$$

For (say) $k=10$ we get: $0.166288$

For (say) $k=10^{20}$ we get: $0.1666666666666666666666666666666666666666625$

How would I prove the limit exists?

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1 Answer 1

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Essentially you have $$f(k) = \left(\sum_{k^2+1}^{k^2+2k} \left( \sqrt{n} - k\right) \right) -k = \sum_{k^2+1}^{k^2+2k}\sqrt{n} -2k \times k - k = \sum_{k^2+1}^{k^2+2k}\sqrt{n} -(2k^2+k)$$ Hence, we need to evaluate $$\sum_{k^2+1}^{k^2+2k}\sqrt{n} = \sum_{1}^{2k}\sqrt{n+k^2} = k \sum_{n=1}^{2k}\sqrt{1+ \dfrac{n}{k^2}}$$ Recall that $$\sqrt{1+x} = 1 + \dfrac12 x + \dfrac{\dfrac12 \times \left(\dfrac12-1\right)}{2!}x^2 + \dfrac{\dfrac12 \times \left(\dfrac12-1\right) \left(\dfrac12-2\right)}{3!}x^3 + \mathcal{O}(x^4)$$ This gives us $$\sum_{n=1}^{2k}\sqrt{1+ \dfrac{n}{k^2}} = \sum_{n=1}^{2k} \left(1 + \dfrac{n}{2k^2} - \dfrac{\dfrac12 \cdot \dfrac12}2 \dfrac{n^2}{k^4} + \mathcal{O} \left(\dfrac{n^3}{k^6}\right)\right) = \sum_{n=1}^{2k} \left(1 + \dfrac{n}{2k^2} - \dfrac{n^2}{8k^4} + \mathcal{O} \left(\dfrac{n^3}{k^6}\right)\right)$$ This gives us $$\sum_{n=1}^{2k}\sqrt{1+ \dfrac{n}{k^2}} = 2k + \dfrac{(2k)(2k+1)}{4k^2}-\dfrac{(2k)(2k+1)(4k+1)}{48k^4} + \mathcal{O} \left(\dfrac1{k^2}\right)\\ = 2k + 1 + \dfrac1{2k} - \dfrac1{3k} + \mathcal{O} \left(\dfrac1{k^2} \right)$$ Hence, $$f(k) = k \sum_{n=1}^{2k}\sqrt{1+ \dfrac{n}{k^2}} - (2k^2+k) = \dfrac16 + \mathcal{O} \left(\dfrac1k\right)$$which gives what you want.

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