# Thread: Limit of a sequence

1. ## Limit of a sequence

Prove that the limit of {Sqrt(n+1)-Sqrt(n)} = 0.

2. Originally Posted by tarheelborn
Prove that the limit of {Sqrt(n+1)-Sqrt(n)} = 0.

Multiply by conjugate: $\sqrt{n+1}-\sqrt{n}=\left(\sqrt{n+1}-\sqrt{n}\right)\,\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{ n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}=$...

Tonio

3. Originally Posted by tarheelborn
Prove that the limit of {Sqrt(n+1)-Sqrt(n)} = 0.
Dear tarheelborn,

You can show this using the limit definition for a sequence.

$i.e:~\lim_{n\rightarrow\infty}\{a_n\}=a\Longleftri ghtarrow~\forall~\epsilon>0~~\exists~n_{o}\in{Z^+} ~such~that~n>n_{o}\Rightarrow{\mid~a_{n}-a~\mid}<\epsilon$

Of course you will have to multiply by the conjugate as tonio had pointed out.

Hope you can continue.

4. Originally Posted by tonio
Multiply by conjugate: $\sqrt{n+1}-\sqrt{n}=\left(\sqrt{n+1}-\sqrt{n}\right)\,\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{ n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}=$...

Tonio
Actually, I got this far on my own. I am not sure how to solve in terms of Epsilon.

5. Would it be acceptable to prove this using 1/sqrt(n+1) instead of the whole denominator 1/(sqrt(n+1)-sqrt(n))?

6. Originally Posted by tarheelborn
not sure how to solve in terms of Epsilon.
Well $\frac{1}{{\sqrt {n + 1} + \sqrt n }} \leqslant \frac{1}
{{2\sqrt n }} < \varepsilon$
.
So how large must n be?

7. So n would be < 1/4*epsilon^2, right?

8. Originally Posted by tarheelborn
So n would be < 1/4*epsilon^2, right?
Try greater than: $n \geqslant \left\lfloor {\frac{1}{{4\varepsilon^2 }} + 1} \right\rfloor$.