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.


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

  • $\begingroup$ Very elegant, exactly what I was looking for. Thanks! $\endgroup$ – 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}

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

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

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