# Thread: Convergence of sequence using the definition

1. ## Convergence of sequence using the definition

Hi, I know how to show the following sequence goes to its limit, but i need to do it using the definition, of which i am having trouble, i've started how i should though i think.

$\displaystyle x_{n} = \sqrt{n+1} - \sqrt{n}$
prove $\displaystyle x_{n} \rightarrow 0$

Let $\displaystyle \epsilon > 0$ be given
Let $\displaystyle N=N(\epsilon )$ be an integer greater than ...(answer goes here)...
Then for all $\displaystyle n>N$ we have
$\displaystyle |x_{n} - l| = |\sqrt{n+1}-\sqrt{n} - 0|$

$\displaystyle |x_{n} - l| = |\sqrt{n+1} - \sqrt{n}|$

$\displaystyle \sqrt{n+1} - \sqrt{n} = \dfrac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}}$
$\displaystyle |x_{n} - l| = \dfrac{1}{\sqrt{n+1} + \sqrt{n}}$

and that is where i get stuck.
I'm not sure if this definition method of proof is well known so if you would like an example of a completed question just ask. Thanks.

2. Originally Posted by Rapid_W
$\displaystyle x_{n} = \sqrt{n+1} - \sqrt{n}$
prove $\displaystyle x_{n} \rightarrow 0$

Let $\displaystyle \epsilon > 0$ be given
Let $\displaystyle N=N(\epsilon )$ be an integer greater than ...(answer goes here)...
Then for all $\displaystyle n>N$ we have
$\displaystyle |x_{n} - l| = |\sqrt{n+1}-\sqrt{n} - 0|$.
The first thing to note is $\displaystyle x_n=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}$.
Then $\displaystyle \left|\dfrac{1}{\sqrt{n+1}+\sqrt{n}} -0\right|\le\dfrac{1}{2\sqrt{n}}$.
Can you finish?

3. Yes, that is the missing link, thanks very much.

i did think of doubling it, but i did this instead $\displaystyle \dfrac{1}{\sqrt{2n}}$, which isn't true lol

Do i have to be able to prove that though, or state a theory?

4. Prove what? That $\displaystyle \sqrt{n+1}- \sqrt{n}= \frac{1}{\sqrt{n+1}+ \sqrt{n}}$?

Rationalize the numerator: multiply both numerator and denominator of $\displaystyle \frac{\sqrt{n+1}- \sqrt{n}}{1}$ by $\displaystyle \sqrt{n+1}+ \sqrt{n}$.

5. no sorry, perhaps i am being silly but, prove that $\displaystyle \dfrac{1}{\sqrt{n+1} + \sqrt{n}} \le \dfrac{1}{2\sqrt{n}}$

also, the answer is $\displaystyle N=N(\epsilon ) = integer > 4\epsilon ^2$?

6. $\displaystyle \sqrt{n+1}+\sqrt{n}\ge\sqrt{n}+\sqrt{n}=2\sqrt{n}$.

SO?

7. so 1 over then all is the same, ok got that, so if i finish of my work i get

$\displaystyle |x_{n}-l| \le \dfrac{1}{2\sqrt{n}}$
$\displaystyle n > N so$
$\displaystyle |x_{n}-l| \le \dfrac{1}{2\sqrt{N}}$
$\displaystyle \dfrac{1}{2\sqrt{N}} < \epsilon$
If $\displaystyle N > 4\epsilon ^2$
$\displaystyle |x_{n}-l| \le \epsilon$

and then i write $\displaystyle 4\epsilon ^2$ in the gap at the top.

Is this right?