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I have been attempting to follow the answer to a question previously asked on the site (Lebesgue integral basics), but am lost on how one might go about evaluating $$\lim_{n\to\infty} \sum\limits_{i=1}^n \frac in\left(\sqrt{(i+1)/n}-\sqrt{i/n}\right)$$ I have been unable to locate a summation formula for the sum of square roots. Any help is greatly appreciated.

Community
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  • Have you tried multiplying by conjugate? – SBF Dec 10 '14 at 08:02
  • No sir. I am quite frankly lost. I have experience with the basic limit techniques taught in an undergraduate calculus series, but it has been some time and this is much more difficult than I remember. –  Dec 10 '14 at 08:17

2 Answers2

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You can manipulate the sums to get this working:

\begin{align} \sum\limits_{i=1}^n\frac{i}{n}(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}})&=\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=\sum\limits_{i=1}^n\frac{i+1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=\sum\limits_{i=2}^{n+1}\frac{i}{n}\sqrt{\frac{i}{n}}-\sum\limits_{i=1}^n\frac{1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=(\frac{n+1}{n})^{\frac{3}{2}}-\sum\limits_{i=1}^n\frac{1}{n}\sqrt{\frac{i+1}{n}}\\&=(\frac{n+1}{n})^{\frac{3}{2}}-(\frac{1}{n})^{3/2}\sum\limits_{i=2}^{n+1}\sqrt{i} \end{align}

Now this is a lot better. I'm going to put bounds on the sum using $\int\sqrt{x}$, because $\sqrt{x}$ is monotone increasing.

$\int\limits_1^{n+1}\sqrt{x}dx\leq\sum\limits_{i=2}^{n+1}\sqrt{i}\leq\int\limits_2^{n+2}\sqrt{x}dx$

$\frac{2}{3}(n+1)^{3/2}-\frac{2}{3}\leq\sum\limits_{i=2}^{n+1}\sqrt{i}\leq\frac{2}{3}(n+2)^{3/2}-\frac{4}{3}\sqrt{2}$

Multiplying through by $(\frac{1}{n})^{3/2}$ gives us:

$\frac{2}{3}(\frac{n+1}{n})^{\frac{3}{2}}-\frac{2}{3n^{\frac{3}{2}}}\leq(\frac{1}{n})^{3/2}\sum\limits_{i=2}^{n+1}\sqrt{i}\leq\frac{2}{3}(\frac{n+2}{n})^{\frac{3}{2}}-\frac{4\sqrt{2}}{3n^{\frac{3}{2}}}$

This directly transfers into a limit inequality:

$\frac{2}{3}\leq\lim\limits_{n\to\infty}(\frac{1}{n})^{\frac{3}{2}}\sum\limits_{i=1}^{n+1}\sqrt{i}\leq\frac{2}{3}$

We see

$\lim\limits_{n\to\infty}(\frac{1}{n})^{\frac{3}{2}}\sum\limits_{i=1}^{n+1}\sqrt{i}=\frac{2}{3}$

Our final limit then becomes

$\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\frac{i}{n}(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}})=1-\frac{2}{3}=\frac{1}{3}$

Benjamin Roycraft
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Let $f(x)=\sqrt{x}$. Then $f'(x)=\frac{1}{2\sqrt{x}}>0$. Let $x_i=\frac{i}{n}$. Thus, by the Mean Value Theorem, we have $$ \sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}}=f(x_{i+1})-f(x_i)=\frac{1}{n}f'(\xi_i)$$ where $\xi_i\in(\frac{i}{n},\frac{i+1}{n})$ and hence \begin{eqnarray} \sum_{i=1}^n\frac{i}{n}(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}})&=&\frac{1}{n}\sum_{i=1}^n\frac{i}{n}f'(\xi_i). \end{eqnarray} But $$ \frac{1}{n}\frac{i}{n}f'(\xi_i)\le\frac{1}{n}\xi_if'(\xi_i)\le \frac{1}{n}\frac{i+1}{n}f'(\xi_i)$$ and hence $$ \frac{1}{n}\sum_{i=1}^n\frac{i}{n}\left(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}}\right)\le\frac{1}{n}\sum_{i=1}^n\xi_if'(\xi_i)\le \frac{1}{n}\sum_{i=1}^n\frac{i+1}{n}\left(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}}\right).$$ Note that $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\xi_if'(\xi_i)=\int_0^1xf'(x)dx=\int_0^1\frac{x}{2\sqrt{x}}dx=\frac{1}{3} $$ and $$\lim_{n\to\infty}\left[\frac{1}{n}\sum_{i=1}^n\frac{i}{n}\left(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}}\right)-\frac{1}{n}\sum_{i=1}^n\frac{i+1}{n}\left(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}}\right)\right]=0$$ and hence $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{i}{n}\left(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}}\right)=\frac{1}{3}. $$

xpaul
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