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Evaluate the limit : $$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)$$

I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area.

Any help would be greatly appreciated, thanks! Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :)

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Thanks for your kind help everyone, but is there any way to show this with simple basic tools that a calculus newbie like myself would recognize? – Adar Hefer Nov 16 '12 at 16:49
up vote 6 down vote accepted

To use as little machinery as possible, observe that $\left(\sqrt n+\frac1{2\sqrt n}\right)^2=n+1+\frac1{4n}>n+1$, hence $$\tag1\sqrt{n+1}<\sqrt n + \frac1{2\sqrt n}.$$ By induction, we see therefore that $$\tag22\sqrt {n+1}<2+\sum_{k=1}^n \frac1{\sqrt k}\quad\text{for all }n\in\mathbb N.$$ On the other hand, if $q>2$, then for $n$ sufficiently large, we have $\left(\sqrt n+\frac1{q\sqrt n}\right)^2=n+\frac2q+\frac1{q^2n}<n+1$ and hence $$\tag3\sqrt{n+1}>\sqrt n + \frac1{q\sqrt n}.$$ Again by induction, we therefore find $$\tag4q\sqrt {n+1}>C+\sum_{k=1}^n \frac1{\sqrt k}\quad\text{for all }n\in\mathbb N,$$ where $C$ is a (negative) constant depending on $q$ (needed to cover the fact that $(3)$ hold only for $n$ sufficiently large). This gives us $$\frac{2\sqrt{n+1}-2}{\sqrt n}<\frac1{\sqrt n}\left(1+\frac1{\sqrt 2}+\cdots+\frac1{\sqrt n}\right)<\frac{q\sqrt{n+1}-C}{\sqrt n}$$ for almost all $n$. The left and right estimate converge to $2$ and $q$, respectively, as $n\to\infty$. Since $q$ was any number $>2$, we conclude that $$\lim_{n\to\infty}\frac1{\sqrt n}\left(1+\frac1{\sqrt 2}+\cdots+\frac1{\sqrt n}\right)=2.$$

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After reading Phira's answer, I suggest you replace the upper bound with something based on $\sqrt{n+1}>\sqrt n+\frac1{2\sqrt {n+1}}$ for all $n$ rather than my $\sqrt{n+1}>\sqrt n+\frac1{q\sqrt {n}}$ for almost all $n$. – Hagen von Eitzen Nov 16 '12 at 17:09
I don;t get why $\left(\sqrt n+\frac1{q\sqrt n}\right)^2=n+\frac q2+\frac1{q^2n}<n+1$ for $q>2$ – Norbert Dec 4 '12 at 15:09
My wrting $\frac q2$ instead of $\frac2q$ was a TeX accident. With correct formula $\left(\sqrt n +\frac 1{\sqrt n}\right)^2 = n+2\cdot\sqrt n\cdot \frac1{q\sqrt n}+\frac1{q^2n}=n+\frac 2q+\frac 1{q\sqrt n}$ and $\frac2q<1$. For sufficiently large $n$, we still have $\frac2q+\frac1{q^2n}<1$. – Hagen von Eitzen Dec 4 '12 at 17:38

Rewrite as $$\frac 1n \left(\frac 1 {\sqrt{\frac 1n}}+\frac 1 {\sqrt{\frac 2n}}+\dots +\frac 1 {\sqrt{\frac nn}} \right)$$ and interpret this as a Riemann sum for the function $$\frac 1{\sqrt x}.$$

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Thanks buddy, but I didn't get to differential and integral mathematics yet. Still battling my way through sequences. – Adar Hefer Nov 16 '12 at 16:45
You would get more useful answers if you gave more context to your question. What tools do you have at your disposal? And write the answer to this question in your original question where it belongs. – Phira Nov 16 '12 at 16:48
Sandwich principle, Cesaro mean theorem, arithmetics with limits.. I'd probably elaborate further but I would need to look up the English terms. Do you figure these are enough, though? – Adar Hefer Nov 16 '12 at 16:53
Does interpreting the sequence as a Riemann sum guarantee convergence to the integral in the general case? I'm not sure it does, since the idea that the integral exists means that we can find a partition that means that the sum is close to the value for the integral. I suppose that in this case we could find n large enough so that there is an $\sqrt{\frac{i}{n}}$ in each subinterval of the given partition and use convexity of the function, but that seems like a lot of hassle, and I'm not sure we can claim it's true for the general case. Am I missing something? – Tom Oldfield Nov 16 '12 at 17:14
@TomOldfield This is a continuous function, thus (Riemann)integrable, thus the Riemann sum converges. What you call "hassle" is part of the general proof that any fine partition works. – Phira Nov 16 '12 at 17:30

Note that $$ 2(\sqrt{k+1}-\sqrt{k})=\frac{2}{\sqrt{k+1}+\sqrt{k}}\leq\frac{1}{\sqrt{k}}\leq \frac{2}{\sqrt{k}+\sqrt{k-1}}=2(\sqrt{k}-\sqrt{k-1}) $$ Hence $$ 2(\sqrt{n+1}-1)=\sum\limits_{k=1}^n 2(\sqrt{k+1}-\sqrt{k})\leq\sum\limits_{k=1}^n\frac{1}{\sqrt{k}}\leq \sum\limits_{k=1}^n 2(\sqrt{k}-\sqrt{k-1})=2\sqrt{n} $$ so $$ \frac{2\sqrt{1 + n}-2}{\sqrt{n}}\leq\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}\leq2 $$ The rest is clear.

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Thanks though, you're brilliant. I didn't get this far in my studies though, and I don't know how to interpret your answer.. – Adar Hefer Nov 16 '12 at 16:40
Norbert, you're the man!! – Adar Hefer Nov 16 '12 at 16:59

This is a standard Stolz-Cezaro problem

$$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)=\lim_{n \to \infty } \frac{\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)}{\sqrt{n}}=$$ $$=\lim_{n}\frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}$$

rationalize the denominator you get

$$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)=\lim_n \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}=2$$

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A couple of other methods, which use more machinery than the earlier ones, but are worth looking into for their generality.

Bounding by integrals

Comparing the integrals by constant bounding function over unit intervals, we get $$ \frac1{\sqrt{n}}\left(\int_1^{n+1}\frac{\mathrm{d}x}{\sqrt{x}}\right) \le\frac1{\sqrt{n}}\sum_{k=1}^n\frac1{\sqrt{k}} \le\frac1{\sqrt{n}}\left(1+\int_1^n\frac{\mathrm{d}x}{\sqrt{x}}\right) $$ which yields $$ \color{#C00000}{\frac1{\sqrt{n}}\left(2\sqrt{n+1}-2\right)} \le\frac1{\sqrt{n}}\sum_{k=1}^n\frac1{\sqrt{k}} \le\color{#C00000}{\frac1{\sqrt{n}}\left(2\sqrt{n}-1\right)} $$ As $n\to\infty$, both bounding terms (in red) tend to $2$. Therefore, by the Squeeze Theorem, we get $$ \lim_{n\to\infty}\frac1{\sqrt{n}}\sum_{k=1}^n\frac1{\sqrt{k}}=2 $$ Euler-Maclaurin Sum Formula

The Euler-Maclaurin Sum Formula says that for some constant $C$, we have $$ \sum_{k=1}^n\frac1{\sqrt k}=2\sqrt{n}+\frac1{2\sqrt n}+C+O\left(n^{-3/2}\right) $$ Dividing by $\sqrt n$ yields $$ \begin{align} \frac1{\sqrt n}\sum_{k=1}^n\frac1{\sqrt k} &=2+\frac C{\sqrt n}+\frac1{2n}+O\left(n^{-2}\right)\\ &\to2 \end{align} $$

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Nice answer robjohn.. – juantheron Nov 3 '13 at 13:35
@juantheron: thanks for the comment. Looking back, I noticed a typo in the second answer. :-) – robjohn Nov 3 '13 at 14:01

Write your expression as ${1\over n}$ times $\biggl(\ldots\biggr)$ and you will see that it can be interpreted as a Riemann sum belonging to a certain integral.

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Thanks for the suggestion, but I need a solution using the sandwich principle, Cesaro means, not a whole lot more advanced than this... I've only just started 4 weeks ago. :/ – Adar Hefer Nov 16 '12 at 16:42

You can also use the identity $$\sqrt {x +1} - \sqrt x = \dfrac 1 {\sqrt{x+1} + \sqrt x }$$

which is between $\dfrac 1{2\sqrt {x+1} }$ and $\dfrac 1 {2\sqrt {x}}$.

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It'd be clearer to add this to your first answer. It is also considered rude, or at least not-so-nice, to write more than one answer to the same question in the same thread. – DonAntonio Nov 16 '12 at 19:30
I suggest that you own your opinion and say that you consider it rude. Furthermore, I suggest that you do so at the recent meta thread dealing with this very topic. – Phira Nov 16 '12 at 22:22
Have it as you want: you can read other cases like this. I was trying to be useful, my bad. – DonAntonio Nov 16 '12 at 22:40

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{Stolz\!-\!Ces\grave{a}ro\ Theorem}$: \begin{align} \color{#f00}{\lim_{n \to \infty}\pars{{1 \over \root{n}}\sum_{k = 1}^{n}{1 \over \root{k}}}} & = \lim_{n \to \infty}{1/\root{n + 1} \over \root{n + 1} - \root{n}} = \lim_{n \to \infty}{\root{n + 1} + \root{n} \over \root{n + 1}} \\[3mm] & = \lim_{n \to \infty}\pars{1 + {1 \over \root{1 + 1/n}}} = \color{#f00}{2} \end{align}

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