# Find the limit of a sequence under given conditions

Let $a_{n(n \geq 1)}$ be a sequence of real numbers such that the sequence $1 + \frac{1}{\sqrt 2} + \cdots + \frac{1}{\sqrt n} - a_n \sqrt n$ is bounded. Find $\lim_{n \rightarrow \infty} a_n$.

Let $b_n = 1 + \frac{1}{\sqrt 2} + \cdots + \frac{1}{\sqrt n} - a_n \sqrt n$.

We observe that $\sqrt n - a_n \sqrt n \leq b_n \leq n - a_n \sqrt n$. Because $b_n$ is bounded, we must have:

$$\lim_{n \rightarrow \infty} \sqrt n - a_n \sqrt n = \lim_{n \rightarrow \infty} \frac{1 - \frac{a_n^2}{n}}{\frac{1}{n} + \frac{a_n}{\sqrt n^3}} \in \mathbb{R}$$ $$\lim_{n \rightarrow \infty} n - a_n \sqrt n \in \mathbb{R} = \lim_{n \rightarrow \infty} \frac{1 - a_n^2}{\sqrt{\frac{1}{n}} + a_n \sqrt{\frac{1}{n}}} \in \mathbb{R}$$

We conclude that $a_n \rightarrow 0$.

Is this a correct approach and is the solution correct?

Thank you!

No. From the fact that $\sqrt{n}-a_n\sqrt{n}$ is bounded above you cannot conclude that it converges; similarly for $n-a_n\sqrt{n}$. Moreover, $\sqrt{n}-a_n\sqrt{n}=(1-a_n)\sqrt{n}$ converges to $\infty$ if $a_n\to0$.

The following is a solution. $$a_n=\frac{1}{\sqrt n}\,\sum_{k=1}^n\frac{1}{\sqrt k}-\frac{b_n}{\sqrt n}.$$ Since $b_n$ is bounded, the second term in the right hand side converges to $0$. The first term can be written as $$\frac{1}{n}\,\sum_{k=1}^n\frac{1}{\sqrt{k/n}},$$ which is a Riemann sum for the improper integral$\int_0^1dx/\sqrt{x}$.

• I haven't studied the notion of Riemann sums yet, so, could you give me a solution without involving this concept? Commented Oct 17, 2016 at 17:45
• Do you know about definite integrals? Commented Oct 18, 2016 at 11:34
• No, I haven't studied them yet. Commented Oct 18, 2016 at 11:45
• I am afraid there is no easy way to avoid the use of integrals (see D'Aurizio's answer.) Commented Oct 18, 2016 at 16:14

We may notice that $$2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}} =\int_{n-1/2}^{n+1/2} \frac{dx}{\sqrt{x}}=\int_{-1/2}^{1/2}\frac{dx}{\sqrt{n+x}}=\frac{1}{\sqrt{n}}-\int_{-1/2}^{1/2}\frac{x\,dx}{\sqrt{n(n+x)}}$$ from which it follows that $$2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n}}+O\left(\frac{1}{n\sqrt{n}}\right)$$ as well as $$\sum_{n=1}^{N}\frac{1}{\sqrt{n}}=O(1)+\sum_{n=1}^{N}\left( 2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}}\right) = 2\sqrt{N}+O(1).$$

• The same result can be obtained without the use of integrals from the Taylor expansion $$\sqrt{1+x}-\sqrt{1-x}=x+O(x^3).$$ Commented Oct 19, 2016 at 14:54
• @JuliánAguirre, of course. Commented Oct 19, 2016 at 15:00