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Find the following limit $:$

$$\lim_{N \to \infty} \frac {1} {\sqrt {N}} \sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}}.$$

It is quite clear that $\sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}} \gt \sqrt {N}$ for $N \gt 1.$ So the limit (if it exists finitely) has to be $\geq 1.$ But I believe that the limit is infinty. For that I need the sum to be greater than some scalar multiple of $N^s$ for sufficiently large $N$ where we require $s \gt \frac {1} {2}.$ Is it possible to attain this lower bound eventually? Any help in this regard would be greatly appreciated.

Thanks a lot.

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    $\begingroup$ The limit is $2$. Use comparison with $\frac 1 {\sqrt N}\int_1^{N} \frac 1 {\sqrt x}dx$. $\endgroup$ Dec 8 '21 at 7:47
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    $\begingroup$ @KaviRamaMurthy$:$ We have the following inequalities $:$ $$\frac {1} {\sqrt {N}} \left (1 + \int_{1}^{N} \frac {1} {\sqrt {x}}\ dx \right ) \geq \frac {1} {\sqrt {N}} \sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}} \geq \frac {1} {\sqrt {N}} \int_{1}^{N} \frac {1} {\sqrt {x}}\ dx.$$ So by Sandwich theorem we have $$\lim\limits_{N \to \infty} \frac {1} {\sqrt {N}} \sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}} = \lim\limits_{N \to \infty} \frac {1} {\sqrt {N}} \int_{1}^{N} \frac {1} {\sqrt {x}}\ dx = \lim\limits_{N \to \infty} \frac {2 \sqrt {N} - 2} {\sqrt {N}} = 2.$$ Am I right? $\endgroup$
    – Fanatics
    Dec 8 '21 at 8:03
  • $\begingroup$ Yes, that is right. $\endgroup$ Dec 8 '21 at 8:05
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    $\begingroup$ You can also consider the Riemann sum $\frac1N\sum_{n=1}^N\frac1{\sqrt\frac nN}$. $\endgroup$
    – nejimban
    Dec 8 '21 at 8:10
  • $\begingroup$ @nejimban$:$ This Riemann sum approximates to the integral $\displaystyle \int_{0}^{1} \frac {1} {\sqrt {x}}\ dx$ for sufficiently large $N.$ $\endgroup$
    – Fanatics
    Dec 8 '21 at 8:16
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I think it will helpful to those who are not familiar with the technique mentioned above in the comment section. So, I am posting an answer.

By definition we know $\int_0^1 f(x) dx=\lim\limits_{N \to\infty} \frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})$.

Also, $$\lim\limits_{N \to\infty} \frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}=\lim\limits_{N \to\infty}\frac{1}{N} \sum_{n=1}^N \sqrt\frac{N}{n}=\lim\limits_{N \to\infty} \frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})$$ where $f(x)=\frac {1}{\sqrt x}$.

Therefore:

$$\lim\limits_{N \to\infty} \frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}=\int_0^1 \frac {1}{\sqrt x}dx=2$$

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  • $\begingroup$ Some comments explaining why the integral converges to the correct sum should be added. Which theorem should be used to conclude that? Note that the function is not continuous at $0$, since $f$ is not defined there. $\endgroup$ Dec 8 '21 at 9:13
  • $\begingroup$ @R.W.Prado yes. Although $f(x)=\frac {1}{\sqrt x}$ is not bounded on $[0,1]$, all the relations still work because $\int_a^1 f(x) dx$ exists as $a$ tends to zero. $\endgroup$
    – Rajaei
    Dec 8 '21 at 9:41
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If you enjoy generalized harmonic numbers, you can have a good appromation os the partial sum $$S_N=\frac {1} {\sqrt {N}} \sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}}=\frac {1} {\sqrt {N}}\,H_N^{\left(\frac{1}{2}\right)}$$ and, using asymptotics $$S_N=2+\frac{\zeta \left(\frac{1}{2}\right)}{\sqrt{N}}+\frac{1}{2 N}-\frac{1}{24 N^2}+O\left(\frac{1}{N^4}\right)$$ Using this trucated series for $N=100$, you would obtain $1.858960382452$ while the exact value is $1.858960382478$

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  • $\begingroup$ More comments should be added explaining how to obtain these numbers and why the sum is not 2, as the others suggests. $\endgroup$ Dec 8 '21 at 9:12
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    $\begingroup$ @R.W.Prado. I computed and approximated the partial sum $S_N$. Now, if $N\to \infty$, the result is $2$. $\endgroup$ Dec 8 '21 at 9:14
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Firstly note that $$2(\sqrt{n+1}-\sqrt{n}) = 2\frac{(\sqrt{n+1}-\sqrt{n}) (\sqrt{n+1}+\sqrt{n}) }{(\sqrt{n+1}+\sqrt{n}) }=\frac{2}{(\sqrt{n+1}+\sqrt{n}) }<\frac{2}{2\sqrt{n}}=\frac{1}{\sqrt{n}}$$ Similarly one can show that : $$2(\sqrt{n}-\sqrt{n-1}) >\frac{1}{\sqrt{n}}$$ Clubbing the inequalities we have $$2{(\sqrt{n+1}-\sqrt{n}) }<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1}) $$ Summing them up from $n=1$ to $n=N$ we get a telescopic sum and hence the inequality becomes : $$ 2(\sqrt{N+1}-1)<\sum \limits_{n=1}^{n=N} \frac{1}{\sqrt{n}}<2(\sqrt{N})$$ Now dividing by $\sqrt{N}$ yields $$\frac{2(\sqrt{N+1}-1)}{\sqrt{N}}<\frac{1}{\sqrt{N}}\sum \limits_{n=1}^{n=N} \frac{1}{\sqrt{n}}<2$$ Letting $N\to \infty $ and using sandwich theorem ,we get the required limit equal to 2. That is :

$$\lim_{N\to \infty} \frac{1}{\sqrt{N}}\sum \limits_{n=1}^{n=N} \frac{1}{\sqrt{n}} =2$$

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Let's use Cesàro-Stolz: $\frac{\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}} - \sum_{k=1}^{n} \frac{1}{\sqrt{k}}}{\sqrt{n+1} - \sqrt{n}} = \frac{1/\sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$ and this obviously converges to $2$ as $n \to \infty$.

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