# Question on evaluation of a limit of a sequence.

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.

• The limit is $2$. Use comparison with $\frac 1 {\sqrt N}\int_1^{N} \frac 1 {\sqrt x}dx$. Dec 8 '21 at 7:47
• @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? Dec 8 '21 at 8:03
• Yes, that is right. Dec 8 '21 at 8:05
• You can also consider the Riemann sum $\frac1N\sum_{n=1}^N\frac1{\sqrt\frac nN}$. Dec 8 '21 at 8:10
• @nejimban$:$ This Riemann sum approximates to the integral $\displaystyle \int_{0}^{1} \frac {1} {\sqrt {x}}\ dx$ for sufficiently large $N.$ Dec 8 '21 at 8:16

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:

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

• 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. Dec 8 '21 at 9:13
• @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. Dec 8 '21 at 9:41

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 $$1.858960382478$$

• More comments should be added explaining how to obtain these numbers and why the sum is not 2, as the others suggests. Dec 8 '21 at 9:12
• @R.W.Prado. I computed and approximated the partial sum $S_N$. Now, if $N\to \infty$, the result is $2$. Dec 8 '21 at 9:14

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 :

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

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 $$n \to \infty$$.