1. ## Testing the convergence

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

$

It is given that $x^2 < 1$

2. 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

$\chi$ $\sigma$

3. ## Could u please explain me

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 ....

4. 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

$\chi$ $\sigma$

5. 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.........