I am trying to find the limit of this infinite sequence:

$$\lim_{n \rightarrow\infty} \frac{1}{n}\left(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+\ldots+1\right)$$

I can see that:

$$\left(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+\ldots+1\right) \lt n$$

So the whole expression is bounded by $1$, but I am having a hard time finding the limit. Any help pointing me into the right direction will be appreciated.

up vote 5 down vote accepted

It is the Riemann Sum that converges to $\displaystyle \int_{0}^1 \sqrt{x}dx$.

  • Very elegant, exactly what I was looking for. Thanks! – amaurs Sep 23 '16 at 21:29

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Another interesting approach is the use of Stolz-Ces$\grave{a}$ro Theorem:

\begin{align} &\color{#f00}{\lim_{n \to \infty}{1 \over n}\pars{\root{1 \over n} + \root{2 \over n} + \root{3 \over n} + \cdots + 1}} = \lim_{n \to \infty}{1 \over n}\sum_{k = 1}^{n}\root{k \over n} = \lim_{n \to \infty}{1 \over n^{3/2}}\sum_{k = 1}^{n}k^{1/2} \\[5mm] = &\ \lim_{n \to \infty}{\pars{n + 1}^{1/2} \over \pars{n + 1}^{3/2} - n^{3/2}} = \lim_{n \to \infty}{\pars{n + 1}^{1/2}\bracks{\pars{n + 1}^{3/2} + n^{3/2}} \over \pars{n + 1}^{3} - n^{3}} \\[5mm] = &\ \lim_{n \to \infty}{\pars{n + 1}^{1/2}\bracks{\pars{n + 1}^{3/2} + n^{3/2}} \over 3n^{2} + 3n + 1} = \color{#f00}{2 \over 3}\lim_{n \to \infty}\bracks{% {\pars{1 + 1/n}^{1/2} \over 1 + 1/n + 1/\pars{3n^{2}}}\,{\pars{1 + 1/n}^{3/2} + 1 \over 2}} = \color{#f00}{2 \over 3} \end{align}

  • Well done! +1!! – Mark Viola Oct 5 '16 at 23:27
  • @Dr.MV $\large ?{\bullet\ \bullet \atop \smile}?$. Thanks. – Felix Marin Oct 6 '16 at 1:07

I thought it might be instructive to present a way forward that does not rely on Riemann sums. To that end, we proceed.

Let the sum $S_n$ be given by

$$S_n=\frac{1}{n^{3/2}}\sum_{k=1}^n \sqrt{k}$$

Now, not that we can bound $S_n$ by the integrals

$$\frac {1}{n^{3/2}}\int_0^n \sqrt{x}\,dx\le \frac{1}{n^{3/2}}\sum_{k=1}^n\le \frac{1}{n^{3/2}}\int_1^{n+1}\sqrt{x}\,dx$$

Application of the squeeze theorem finishes the trick.

  • Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Oct 5 '16 at 23:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.