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What's the the result of:

$$\sum_{k=1}^{n}{\sqrt{k}+1}$$

Thanks.

Martin Sleziak
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Santiago Gil
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  • I think there is no closed form, but you can find an asymptotic expansion easily with an integral comparaison – M.LTA Oct 18 '15 at 10:13
  • Why do you expect a closed form of this series? – Dontknowanything Oct 18 '15 at 10:17
  • http://ramanujan.sirinudi.org/Volumes/published/ram09.pdf. This question has been asked before, I cant find it. – Aditya Agarwal Oct 18 '15 at 10:19
  • http://arxiv.org/pdf/1204.0877.pdf – Aditya Agarwal Oct 18 '15 at 10:19
  • I expect as @M.LTA says an integral comparison. I know that it's divergent, but want an approximation (for example to know sum to n=100, n=200...) – Santiago Gil Oct 18 '15 at 10:20
  • If you just want the sum for a few values of $n$, you write a few lines of code to evaluate it. – Gerry Myerson Oct 18 '15 at 10:34
  • For asymptotic behavior of the sum, see answers of this question. This give $\sum_{k=1}^n\left(\sqrt{k}+1\right) \approx \frac23 n^{3/2} + n + \frac12 n^{1/2} + \zeta(-\frac12) + \frac{1}{24}n^{-1/2} + O(n^{-3/2})$. For $n = 200$, this expression matches the exact value with an error $\sim 10^{-9}$. – achille hui Oct 18 '15 at 10:34
  • Sine related posts: http://math.stackexchange.com/questions/422559/what-is-sum-limits-i-1n-sqrt-i, http://math.stackexchange.com/questions/1431489/estimating-sum-n-1-2 and http://math.stackexchange.com/questions/79115/limit-lim-limits-n-rightarrow-infty-left2-sqrt-n-sum-limits-k-1n-frac (The first one seems like a reasonable duplicate targed to me.) – Martin Sleziak Oct 18 '15 at 11:20

1 Answers1

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I will assume we want an approximation of $S_n=\sum_{k=1}^n \sqrt{k}$.

First, observe that $\sqrt{k} \leqslant \int_{k}^{k+1} \sqrt{x} \; \mathrm{d}x \leqslant \sqrt{k+1}$ so that : $$ \int_{k-1}^{k} \sqrt{x} \; \mathrm{d}x \leqslant \sqrt{k} \leqslant \int_{k}^{k+1} \sqrt{x} \; \mathrm{d}x $$ by summation we have : $$\int_{0}^{n} \sqrt{x} \; \mathrm{d}x \leqslant \sum_{k=1}^{n+1} \sqrt{k} \leqslant \int_{1}^n \sqrt{x} \; \mathrm{d}x$$ wich lead to :

$$\frac{2n^{3/2}}{3} \leqslant \sum_{k=1}^{n+1} \sqrt{k} \leqslant \frac{2\left((n+1)^{3/2}-1\right)}{3} $$

You can continue further if you want, always the same method.

M.LTA
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  • Thanks, it's what I searched. Is there another approximation to k = i (this means, the serie doesn't begin with 1). – Santiago Gil Oct 18 '15 at 10:53
  • Yes of course, you just take the summation from i to n (when I said "by summation we have") – M.LTA Oct 18 '15 at 10:56