Testing the convergence

**zorro** · May 3rd 2009, 10:16 PM

Test the convergence of the series
$<br /> \frac{1}{2\sqrt{1}} + \frac{x^2}{3\sqrt{2}} + \frac{x^4}{4\sqrt{3}}+\frac{x^6}{5\sqrt{4}}+.... \infty<br /> <br />$

It is given that $x^2 < 1$

**chisigma** · May 4th 2009, 07:23 AM

The series is...

$\sum_{n=0}^{\infty} a_{n}\cdot x^{2n}= \sum_{n=0}^{\infty}\frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}}$ (1)

Because is $a_{n}<1$ , $\forall n$ and the series...

$\sum_{n=0}^{\infty} x^{2n}$ (2)

... converges for $|x|<1$ , the series (1) and (2) have the same radious of convergence...

Kind regards

$\chi$ $\sigma$

**zorro** · December 14th 2009, 01:25 PM

Originally Posted by chisigma

The series is...

$\sum_{n=0}^{\infty} a_{n}\cdot x^{2n}= \sum_{n=0}^{\infty}\frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}}$ (1)

Because is $a_{n}<1$ , $\forall n$ and the series...

$\sum_{n=0}^{\infty} x^{2n}$ (2)

... converges for $|x|<1$ , the series (1) and (2) have the same radious of convergence...

Kind regards

$\chi$ $\sigma$

thanks for ur reply but i couldnt understand it could u please explain me ....

**chisigma** · December 14th 2009, 11:32 PM

In this case we can prove the convergence of the series...

$\sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}}$ (1)

... using the 'comparison test'...

Convergence tests - Wikipedia, the free encyclopedia

The series to be compared are...

$\sum_{n=0}^{\infty} a_{n}$ , $a_{n}= \frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}}$

$\sum_{n=0}^{\infty} b_{n}$ , $b_{n}= x^{2n}$ (2)

The series of the $b_{n}= x^{2n}$ is 'geometric' and it converges for $|x|<1$ . Since $\forall n$ is $|a_{n}|<|b_{n}|$ the convergence test extablishes that the (1) converges also for $|x|<1$ ...

Merry Christmas from Italy

$\chi$ $\sigma$

**zorro** · December 23rd 2009, 01:23 AM

Originally Posted by chisigma

The series is...

$\sum_{n=0}^{\infty} a_{n}\cdot x^{2n}= \sum_{n=0}^{\infty}\frac{x^{2n}}{(n+2)\cdot \sqrt{n+1}}$ (1)

Because is $a_{n}<1$ , $\forall n$ and the series...

$\sum_{n=0}^{\infty} x^{2n}$ (2)

... converges for $|x|<1$ , the series (1) and (2) have the same radious of convergence...

Kind regards

$\chi$ $\sigma$

what is $a_n$ ............i didnt get that part.........

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Testing the convergence

Could u please explain me

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