Proving that the limit of the following sequence $\frac {\sqrt {n}}{1 + \sqrt{n}}$ when $n \rightarrow \infty$ is 1.

Question

Proving that the limit of the following sequence:

$$\lim_{n \to \infty} \frac {\sqrt {n}}{1 + \sqrt{n}} = 1$$

I have selected the $N$ to be $\lfloor (\frac{1}{\epsilon} - 1)^2 \rfloor + 1$, am I right?

Answer 1

We have $$ \left|\frac{\sqrt{n}}{1+\sqrt{n}} - 1\right| = \frac{1}{1+\sqrt{n}} , $$ so we want $$ \frac{1}{1+\sqrt n} < \epsilon \iff \sqrt{n} > \frac{1}{\epsilon}-1 \iff n > \left( \frac{1}{\epsilon}-1 \right)^2, $$ so you can indeed take $$ N = \left\lfloor \left( \frac{1}{\epsilon}-1 \right)^2 \right\rfloor + 1. $$

Answer 2

I think you have it.

We know $0 < \sqrt n < 1+ \sqrt n$ so $\frac {\sqrt n}{1 + \sqrt n} <1$ so we need to prove that for any $\epsilon > 0$ then there is an $N$ so the $n > N$ would imply

$0< 1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon$

And $1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon \iff$

$(1 + \sqrt n) - \sqrt n= 1 < \epsilon(1+\sqrt n) \iff $

$\frac 1{\epsilon} -1 < \sqrt n$. Wolog we may assume $\epsilon < 1$. If we prove it is true for any $0 < \epsilon_1 < 1$ then the same $N$ will work for any larger $\epsilon \ge \epsilon_1$. So if we assumme $0 < \epsilon < 1$ then $\frac 1{\epsilon} > 1$ and

$\frac 1{\epsilon} -1 < \sqrt n \iff$

$(\frac 1{\epsilon} -1)^2 < n$ so

So for any $N \ge (\frac 1{\epsilon} -1)^2$ we will have that if $n > N$ then $0< 1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon$.

So, ... Okay, I'd have picked the ceiling of $(\frac 1{\epsilon} -1)^2$ and there's no brownie points in figuring the exact minimum $N$ we can choose.

But yes if $N = \lfloor (\frac 1{\epsilon} -1)^2 \rfloor < n$ and $n \in \mathbb N$ then $n \ge N+1 > (\frac 1{\epsilon} -1)^2$ and therefore $0< 1 - \frac {\sqrt n}{1 + \sqrt n} < \epsilon$.

So yes. Using the floor will work and is correct. And so will any $N$ greater than the floor.

Actually, is there any requirement in your definition that $N$ must actually be a natural number? There is no logistical requirement for this if you think about it. After all it is $n > N$ that must be natural. The $N$ is just a hurdle and there's no reason that can't just be some real magnitude.

So actually, I'd just choose $N = (\frac 1{\epsilon} -1)^2$ and be done with.

Answer 3

As

$$\frac{\sqrt n}{1+\sqrt n}-1=\frac1{1+\sqrt n},$$ which is a decreasing function, you can work this out by solving the equation

$$\frac1{1+\sqrt x}=\epsilon,$$ the solution of which is obviously

$$x=\left(\frac1\epsilon-1\right)^2.$$

Now you want the first integer $n$ strictly larger than $x$, and this is $\lfloor x\rfloor+1$.

Answer 4

Let $f(n)$ be a increasing positive sequence such that, for any $c > 0$, there is an $N(c)$ such that $f(N(c)) > c$.

Then, for any reals $a$ and $b$, $\lim_{n \to \infty} \dfrac{f(n)+a}{f(n)+b} =1$.

Proof.

$\dfrac{f(n)+a}{f(n)+b}-1 =\dfrac{(f(n)+a)-(f(n)+b)}{f(n)+b} =\dfrac{a-b}{f(n)+b} $.

For any $\epsilon > 0$ we want an $n$ such that $|\dfrac{a-b}{f(n)+b}| \le \epsilon$.

This will be true if $\dfrac{|a-b|}{f(n)-|b|} \le \epsilon $ or $|a-b| \le \epsilon(f(n)-|b|) $ or $f(n) \ge \dfrac{|a-b|}{\epsilon}+|b| $.

Therefore, if $n > N(\dfrac{|a-b|}{\epsilon}+|b|) $, $|\dfrac{f(n)+a}{f(n)+b}-1| \lt \epsilon $, so $\lim_{n \to \infty} \dfrac{f(n)+a}{f(n)+b} = 1$.

Now let $a=0, b=1, f(n) = \sqrt{n}$. We can choose $N(c) = c^2+1$.