# limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$

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

• I think you can now accept an answer. You have been given a great solution by Dane. – Pedro Tamaroff Feb 26 '12 at 17:19
• This limit was evaluated at this MSE link. – Marko Riedel Feb 15 '14 at 0:33

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