Limit of a sequence

**tarheelborn** · June 7th 2010, 05:16 PM

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

**tonio** · June 7th 2010, 06:11 PM

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

**Sudharaka** · June 7th 2010, 06:20 PM

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.

**tarheelborn** · June 8th 2010, 05:10 AM

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.

**tarheelborn** · June 8th 2010, 06:24 AM

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

**Plato** · June 8th 2010, 06:26 AM

Originally Posted by tarheelborn

not sure how to solve in terms of Epsilon.

Well $\frac{1}{{\sqrt {n + 1} + \sqrt n }} \leqslant \frac{1}<br /> {{2\sqrt n }} < \varepsilon$ .
So how large must n be?

**tarheelborn** · June 8th 2010, 06:28 AM

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

**Plato** · June 8th 2010, 06:32 AM

Originally Posted by tarheelborn

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

Try greater than: $n \geqslant \left\lfloor {\frac{1}{{4\varepsilon^2 }} + 1} \right\rfloor$ .

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