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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?2012-11-16

7 Answers 7

3

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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    Have it as you want: you can read other cases like this. I was trying to be useful, my bad.2012-11-16
20

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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    Norbert, you're the man!!2012-11-16
18

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 $\frac 1{\sqrt x}.$

  • 0
    Actually, nevermind, I've looked up some things now and it seems that the definition of Riemann integrable that I've been given actually means Darboux integrable, although the two conditions are actually the same (apparently). What I was asking then, boils down to asking why a function that is Darboux integrable is Riemann integrable, and this is the statement I wasn't sure was true (and then fallaciously thought was wrong).2012-11-16
13

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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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} 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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    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}<1.2012-12-04
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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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    @juantheron: thanks for the comment. Looking back, I noticed a typo in the second answer. :-)2013-11-03
4

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. :/2012-11-16