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Calculate below limit $$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$

jimjim
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qwerty89
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2 Answers2

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As a consequence of Euler's Summation Formula, for $s > 0$, $s \neq 1$ we have $$ \sum_{j =1}^n \frac{1}{j^s} = \frac{n^{1-s}}{1-s} + \zeta(s) + O(|n^{-s}|), $$ where $\zeta$ is the Riemann zeta function. In your situation, $s=1/2$, so $$ \sum_{j =1}^n \frac{1}{\sqrt{j}} = 2\sqrt{n} + \zeta(1/2) + O(n^{-1/2}) , $$ and we have the limit $$ \lim_{n\to \infty} \left( \sum_{j =1}^n \frac{1}{\sqrt{j}} - 2\sqrt{n} \right) = \lim_{n\to \infty} \big( \zeta(1/2) + O(n^{-1/2}) \big) = \zeta(1/2). $$

Greg Martin
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Cardboard Box
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    Little problem with your argument : $\zeta(1/2)$ is either infinite (if you don't consider the extension of $\zeta$ to the complex plane without the pole) or finite (if you consider it). But the finite sum where $\zeta$ appears is only consistent if $\zeta(1/2)$ is finite, and your result in the end only works if $\zeta(1/2)$ is infinite (see JavaMan's answer below). So there's a bit of inconsistency somewhere in this answer. And I think there's a typo in your first line : you should replace $1/n^s$ by $1/i^s$. – Patrick Da Silva Feb 15 '12 at 16:50
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    @PatrickDaSilva: no, Dane's first formula holds even though the series on the left doesn't converge as $n\to\infty$. (In fact, the first term on the right-hand side indicates exactly how the series diverges.) Check chapter 1 of Montgomery-Vaughan's book or other books on analytic number theory - or do the partial summation argument yourself. – Greg Martin Feb 15 '12 at 16:54
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    @Greg : So letting $n=1$ you're saying that $1/1 = 1^{1/2}/(1-1/2) + \zeta(1/2) + O(1^{-1/2})$ makes sense? This is saying that $1 = \infty + O(1)$ ; I certainly don't agree with that. – Patrick Da Silva Feb 15 '12 at 16:57
  • @Greg: Damn, I am so wrong. I computed things too quick and forgot to subtract $2 \sqrt n$ while following JavaMan's hint. I said nothing. – Patrick Da Silva Feb 15 '12 at 17:03
  • @PatrickDaSilva: when you say that $\zeta(1/2)=\infty$, you're mistaken. $\zeta(1/2)$ is a well-defined number that comes from the analytic continuation of $\zeta$; the divergence of the series $\sum j^{-1/2}$ is irrelevant. – Greg Martin Feb 15 '12 at 17:03
  • +1 Just trying to calculate this and extrapolating convergence, I produced $-1.46035450881155$. This is subject to rounding errors but $\zeta(1/2) \approx -1.4603545088095868$ so only the last three or four digits of my calculations were wrong. – Henry Feb 15 '12 at 17:18
  • @PatrickDaSilva: Sorry about that. I realized my asymptotics were far too imprecise to be useful, so I deleted my answer. – JavaMan Feb 15 '12 at 18:50
  • To JavaMan : I know that. @Greg : I know the analytic continuation thing, and at first that was precisely my point ; since I was confused with JavaMan's hint that I mistakenly followed (my mistake, not his), I thought the limit was worth infinity ; when Dane wrote that the limit was $\zeta(1/2)$, I wondered if he meant $\sum_{n=1}^{\infty} 1/(n^{1/2}) = \infty$ or $\zeta(1/2)$. – Patrick Da Silva Feb 15 '12 at 20:23
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    (+1) In this answer, it is shown, using analytic continuation, that $\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^nk^{-z}-\frac1{1-z}n^{1-z}\right)=\zeta(z)$ for $0\lt\mathrm{Re}(z)\lt1$. – robjohn Oct 14 '20 at 05:29
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The following is an elementary consideration, which shows how to compute the limit in terms of an infinite series. It's evaluation requires usage of Euler's summation formula, already covered by Dane.

Consider the following transformation $$ \sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \left(\frac{1}{\sqrt{k}} - \frac{2}{\sqrt{k} + \sqrt{k+1}} \right) + \sum_{k=1}^n \frac{2}{\sqrt{k} + \sqrt{k+1}} $$ Then use $\sqrt{k+1}-\sqrt{k} = \frac{\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}} = \frac{(k+1)-k}{\sqrt{k+1}+\sqrt{k}} = \frac{1}{\sqrt{k+1}+\sqrt{k}}$: $$ \sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right) $$ The latter sum telescopes: $$ \sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right) = \left( \sqrt{2}-\sqrt{1} \right) + \left( \sqrt{3}-\sqrt{2} \right) + \cdots + \left( \sqrt{n+1}-\sqrt{n} \right) = \sqrt{n+1}-1 $$ From here: $$ \begin{eqnarray} \left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} &=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \left( \sqrt{n+1}-\sqrt{n}-1\right) \\ &=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \left( \frac{1}{\sqrt{n+1}+\sqrt{n}}-1\right) \end{eqnarray} $$ In the limit: $$ \lim_{n\to \infty} \left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} = -2 + \sum_{k=1}^\infty \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} $$

Patrick Da Silva
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Sasha
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