Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 7 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.