# Convergence of sequence using the definition

• Aug 2nd 2010, 05:55 AM
Rapid_W
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.

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

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

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

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

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.
• Aug 2nd 2010, 07:25 AM
Plato
Quote:

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

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

The first thing to note is $x_n=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}$.
Then $\left|\dfrac{1}{\sqrt{n+1}+\sqrt{n}} -0\right|\le\dfrac{1}{2\sqrt{n}}$.
Can you finish?
• Aug 2nd 2010, 07:32 AM
Rapid_W
Yes, that is the missing link, thanks very much.

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

Do i have to be able to prove that though, or state a theory?
• Aug 2nd 2010, 07:35 AM
HallsofIvy
Prove what? That $\sqrt{n+1}- \sqrt{n}= \frac{1}{\sqrt{n+1}+ \sqrt{n}}$?

Rationalize the numerator: multiply both numerator and denominator of $\frac{\sqrt{n+1}- \sqrt{n}}{1}$ by $\sqrt{n+1}+ \sqrt{n}$.
• Aug 2nd 2010, 07:41 AM
Rapid_W
no sorry, perhaps i am being silly but, prove that $\dfrac{1}{\sqrt{n+1} + \sqrt{n}} \le \dfrac{1}{2\sqrt{n}}$

also, the answer is $N=N(\epsilon ) = integer > 4\epsilon ^2$?
• Aug 2nd 2010, 08:11 AM
Plato
$\sqrt{n+1}+\sqrt{n}\ge\sqrt{n}+\sqrt{n}=2\sqrt{n}$.

SO?
• Aug 2nd 2010, 08:47 AM
Rapid_W
so 1 over then all is the same, ok got that, so if i finish of my work i get

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

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

Is this right?