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Evaluate $$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}.$$

I am trying to use the Sandwich principle here..

$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}=\lim_{n \to \infty}\dfrac{\sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+\cdots+\sqrt{\dfrac{n-1}{n}}}{n}\ge \lim_{n \to \infty}\bigg(\sqrt{\dfrac{1}{n}}.\sqrt{\dfrac{2}{n}}.\sqrt{\dfrac{3}{n}}\cdots\sqrt{\dfrac{n-1}{n}}\bigg )^\dfrac{1}{n-1}=\lim_{n \to \infty}\bigg(\dfrac{(n-1)!}{n^n}\bigg )^\dfrac{1}{2(n-1)}$

But after this I am a little in doubt. This link may provide some light Evaluation of the limit $\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)$ but I do not understand how it would help my problem..

In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link

$\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)=\lim\limits_{n \to \infty }\sum_{k=1}^{n}{\dfrac{1}{\sqrt{kn}}}=\int_{0}^{1}\dfrac{1}{\sqrt{x}}dx+\lim\limits_{n \to \infty }\dfrac{1}{n}=2$

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    $\begingroup$ Well, this is just a Riemann sum. $\endgroup$ – カカロット Dec 15 '18 at 8:45
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    $\begingroup$ Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry $\endgroup$ – Saradamani Dec 15 '18 at 8:46
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    $\begingroup$ Okay then :D Also post it as answer when you finish solving it. $\endgroup$ – カカロット Dec 15 '18 at 8:48
  • $\begingroup$ There are $n-1$ terms in the numerator $\endgroup$ – Shubham Johri Dec 15 '18 at 8:51
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Hint :

$$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n-1}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} \sqrt{\frac{k}{n}} = \int_0^1\sqrt{x}\mathrm{d}x$$

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Just for your curiosity since you already received good answers.

We can approximate the value of $$a_n=\frac{\sum_{i=1}^{n-1} \sqrt i }{n \sqrt n}$$ $$\sum_{i=1}^{n-1} \sqrt i =\sum_{i=1}^{n} \sqrt i -\sqrt n=H_n^{\left(-\frac{1}{2}\right)}-\sqrt n$$ where appear generalized harmonic numbers.

Now, using the asymptotics $$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n\sqrt n}{3}+\frac{\sqrt{n}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt n}+O\left(\frac{1}{n^{5/2}} \right)$$ making, for large $n$ $$a_n=\frac{\frac{2 n\sqrt n}{3}-\frac{\sqrt{n}}{2}+\zeta \left(-\frac{1}{2}\right)+\frac{1}{24\sqrt n}+O\left(\frac{1}{n^{5/2}} \right)} {n \sqrt n}=\frac{2}{3}-\frac{1}{2 n}+\frac{\zeta \left(-\frac{1}{2}\right) } {n \sqrt n }+O\left(\frac{1}{n^{2}} \right)$$ which, for sure, shows the limit and also how it is approached.

But it also gives a good approximation even for small values of $n$. Using $\zeta \left(-\frac{1}{2}\right)\approx -0.207886$ and $n=10$, this would give $a_{10}\approx 0.610093$ while the "exact" value is $a_{10}\approx 0.610509$.

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  • $\begingroup$ It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-\dfrac{1}{2}}$. That's something I have known only today. $\endgroup$ – Saradamani Dec 16 '18 at 5:56
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One more approach. By Stolz-Cesaro Theorem $$\begin{align}\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n-1}}{n\sqrt{n}} &= \lim_{n \to \infty} \frac{\sqrt{n}}{(n+1)\sqrt{n+1}-n\sqrt{n}} \\&=\lim_{n \to \infty} \frac{1}{(n+1)\sqrt{1+\frac{1}{n}}-n}\\ &=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\end{align}$$ where we used the fact that $\sqrt{1+\frac{1}{n}}=1+\frac{1}{2n}+o(1/n)$.

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  • $\begingroup$ But I do not understand how this Stolz-Cesaro Theorem would help me get the $\dfrac{\sqrt{n}}{(n+1){\sqrt{n+1}}-n\sqrt{n}}$ .Please elaborate because according to this theorem if $\lim_{n \to \infty}\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l \implies \lim_{n \to \infty} \dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this.. $\endgroup$ – Saradamani Dec 15 '18 at 16:35
  • $\begingroup$ @Saradamani I added a few details. Is it clear now? $\endgroup$ – Robert Z Dec 15 '18 at 17:29
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    $\begingroup$ @Saradamani For Stolz-Cesaro we have $a_n=\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n-1}$ and $b_n=n\sqrt{n}$. $\endgroup$ – Robert Z Dec 15 '18 at 17:31
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    $\begingroup$ No, If $a_n=\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n-1}$ then $a_{n+1}-a_n=(\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n-1}+\sqrt{n})-(\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n-1})=\sqrt{n}$ $\endgroup$ – Robert Z Dec 16 '18 at 6:38
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    $\begingroup$ @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle... $\endgroup$ – Robert Z Dec 16 '18 at 7:05
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It is sufficient to obtain integral of $\sqrt{x}$, when $0\leq x\leq 1$.

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