Need help on showing convergence of sqrt(n+1)-sqrt(n)

**Skerven** · October 20th 2009, 07:33 AM

y_n := sqrt(n+1) - sqrt(n)

the question asks us to show that y_n and sqrt(n)*y_n converge and to find their limits

i can see that y_n would converge to 0, but i can't show how. i don't even know what the second converges to.

my strategy was to use the x_(n+1)/x_n theorem, but I was left with a messy equation whose comparison to 1 i can't illustrate.

at the very least i ask what theorem i should use. thanks.

**tonio** · October 20th 2009, 07:45 AM

Originally Posted by Skerven

y_n := sqrt(n+1) - sqrt(n)

the question asks us to show that y_n and sqrt(n)*y_n converge and to find their limits

i can see that y_n would converge to 0, but i can't show how. i don't even know what the second converges to.

my strategy was to use the x_(n+1)/x_n theorem, but I was left with a messy equation whose comparison to 1 i can't illustrate.

at the very least i ask what theorem i should use. thanks.

Multiply the expression by 1:

$\left(\sqrt{n+1}-\sqrt{n}\right)\left(\frac{\sqrt{n+1}+\sqrt{n}}{\s qrt{n+1}+\sqrt{n}}\right)$

Develop the above and get that the limit indeed is zero.

Tonio

**Skerven** · October 20th 2009, 07:49 AM

Originally Posted by tonio

Multiply the expression by 1:

$\left(\sqrt{n+1}-\sqrt{n}\right)\left(\frac{\sqrt{n+1}+\sqrt{n}}{\s qrt{n+1}+\sqrt{n}}\right)$

Develop the above and get that the limit indeed is zero.

Tonio

Ok I tried the conjugate technique but I got a messy equation. Ok I'll do it again thanks..

**chisigma** · October 20th 2009, 07:49 AM

The following identity may be useful for You...

$\sqrt{n+1} - \sqrt{n} = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}= \frac{1}{\sqrt{n+1}+\sqrt{n}}$

Kind regards

$\chi$ $\sigma$

**Skerven** · October 20th 2009, 08:04 AM

Originally Posted by chisigma

The following identity may be useful for You...

$\sqrt{n+1} - \sqrt{n} = \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}= \frac{1}{\sqrt{n+1}+\sqrt{n}}$

Kind regards

$\chi$ $\sigma$

how will this help me in finding the limit of sqrt(n)*y_n?

*edit* sorry guys i solved it myself sqrt(n)*y_n --> 1/2. You just reduce the num. and the denom. to convergent sequences through alg. manipulation and apply a theorem or two.

**chisigma** · October 20th 2009, 10:22 AM

Is...

$\lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} = \lim_{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{2}$

Kind regards

$\chi$ $\sigma$

**tonio** · October 20th 2009, 10:43 AM

Originally Posted by Skerven

how will this help me in finding the limit of sqrt(n)*y_n?

*edit* sorry guys i solved it myself sqrt(n)*y_n --> 1/2. You just reduce the num. and the denom. to convergent sequences through alg. manipulation and apply a theorem or two.

Yes, of course...but you could NOT do that before you simplified the expression, right? We already knew the limit is 1/2, you didn't.

Tonio

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