1. ## Convergence Sequence (Analysis)

Hi, im not sure if this is in calculus or number theory section...

Using the definition of a convergent sequence show that $
\sqrt{n+1} - \sqrt{n}
$
converges to zero.

-ynot-

2. Originally Posted by ynotidas
Hi, im not sure if this is in calculus or number theory section...

Using the definition of a convergent sequence show that $
\sqrt{n+1} - \sqrt{n}
$
converges to zero.

-ynot-
Big Hint: multiply by the conjugate over itself and simplify. of course, you have to do this within the framework of the definition

3. Thanks!!
i got to

$1/(\sqrt{n+1} + \sqrt{n})$

i have no idea what to do next??

4. Originally Posted by ynotidas
Thanks!!
i got to

$1/(\sqrt{n+1} + \sqrt{n})$

i have no idea what to do next??
ok, another hint:

$\Bigg| \frac 1{\sqrt{n + 1} + \sqrt{n} }\Bigg| \le \Bigg| \frac 1{\sqrt{n} + \sqrt{n} } \Bigg| = \Bigg| \frac 1{2 \sqrt{n} }\Bigg|$

now see if you can finish up. look at the definition of convergence to see what you have to show

5. thanks, do i do this?

$\frac 1{2\sqrt{n}}

then n>1/(4epsilon^2)

thus |An - 0| < epsilon if n>1 & n>1/(4epsilon^2)

n*> max{1,1/(4epsilon^2)}

n>=n*, we have |An-0|<epsilon

[SORRY it's taking me forever to type it in the [math} mode]

6. Originally Posted by ynotidas
thanks, do i do this?

$\frac 1{2\sqrt{n}}

then n>1/(4epsilon^2)

thus |An - 0| < epsilon if n>1 & n>1/(4epsilon^2)

n*> max{1,1/(4epsilon^2)}

n>=n*, we have |An-0|<epsilon

[SORRY it's taking me forever to type it in the [math} mode]
you want to show that

For every $\epsilon > 0$, there exists an $N \in \mathbb{N}$, such that $n > N$ implies

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

So let $\epsilon > 0$ and choose $N$ such that $N > \frac 1{4 \epsilon ^2}$

Then blah blah blah $< \epsilon$

QED