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

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    I think you can now accept an answer. You have been given a great solution by Dane.2012-02-26
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    This limit was evaluated at this [MSE link](http://math.stackexchange.com/questions/79115/limit-lim-limits-n-rightarrow-infty-left2-sqrt-n-sum-limits-k-1n-frac).2014-02-15

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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 $$ \lim_{n\to \infty} \left( \sum_{j =1}^n \frac{1}{\sqrt{j}} - 2\sqrt{n} \right) = \lim_{n\to \infty} \big( \zeta(1/2) + O(n^{-1/2}) \big) = \zeta(1/2). $$

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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 $1/i^s$.2012-02-15
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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.2012-02-15
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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.2012-02-15
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    @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.2012-02-15
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    @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.2012-02-15
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    +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.2012-02-15
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    @PatrickDaSilva: Sorry about that. I realized my asymptotics were far too imprecise to be useful, so I deleted my answer.2012-02-15
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    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 $\zeta(1/2)$.2012-02-15
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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: $$ \lim_{n\to \infty} \left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} = -2 + \sum_{k=1}^\infty \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} $$