Question on evaluation of a limit of a sequence.

Question

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.

Answer 1

I think it will be helpful to those who are not familiar with the technique mentioned above in the comment section. So, I am posting an answer.

By the definition of Riemann integrals, 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:

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

---

Update:

As mentioned in the comments, $f(x)=\frac{1}{\sqrt x}$ is not Riemann integrable on $[0,1]$, so some explanation is required to justify the last step. Indeed, we should be more careful with the approximation sum.

Note that $f(x)=\frac{1}{\sqrt x}$ is Riemann integrable on $[a,1+a]$ for $a>0$, and in particular, let's assume $2-\epsilon<2\sqrt{1+a}-2\sqrt a$, where $\epsilon$ is an arbitrary positive real number. Hence, by the definition of Riemann integrals, we have:

$$2\sqrt{1+a}-2\sqrt a=\int_a^{1+a} \frac{1}{\sqrt x} dx=\int_a^{1+a} f(x) dx=\lim\limits_{N \to\infty} \frac{1}{N}\sum_{n=1}^N f (a+\frac{n}{N}).$$ So, by the definition of limit, there is $N_0 \in \mathbb N$, such that for every $N>N_0$, we have:

$$2\sqrt{1+a}-2\sqrt a-\epsilon <\frac{1}{N}\sum_{n=1}^N f (a+\frac{n}{N})<2\sqrt{1+a}-2\sqrt a+\epsilon.$$

But for every $N$, we also have:

$$\frac{1}{N}\sum_{n=1}^N f (a+\frac{n}{N})<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<\frac{1}{\sqrt N}\sum_{n=1}^N 2(\sqrt n-\sqrt{n-1})=2.$$

So, for $N>N_0:$

$$2\sqrt{1+a}-2\sqrt a-\epsilon<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<2,$$

and:

$$2-\epsilon -\epsilon <2\sqrt{1+a}-2\sqrt a-\epsilon<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<2.$$

Answer 2

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 $1.858960382478$

Answer 3

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 $n \to \infty$.

Answer 4

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 :

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