# Thread: Convergence Problem

1. ## Convergence Problem

This is the equation that i have ended up with

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

I am using the Ratio test $\frac{u_{n+1} }{u_n} = k$ ....if $k<1 (convergent)$ $k>1(divergent)$ ....

$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\lim_{n \to \infty} \frac{x^2}{n+3 \sqrt{n+2}} . \frac{n+2 \sqrt{n+1}}{1}$

= $x^2 \frac{n+2 \sqrt{n+1}}{n+3 \sqrt{n+2}}$ .........I am stuck here what should i do next?????

2. Your constants are off and you need a ().
Then let n go to infinity and the limit will be $x^2$
set that less than 1 and you will see that in the interval of convergence is (-1,1).
THEN test the endpoint and since the series will converge at both endpoints, the interval of convergence is [-1,1].
PLUS you should rewrite the n's as $(n+1)^{3/2}$
And this is just calc 2.

3. Originally Posted by matheagle
Your constants are off and you need a ().
Then let n go to infinity and the limit will be $x^2$
set that less than 1 and you will see that in the interval of convergence is (-1,1).
THEN test the endpoint and since the series will converge at both endpoints, the interval of convergence is [-1,1].
PLUS you should rewrite the n's as $(n+1)^{3/2}$
And this is just calc 2.
I didnt understand u mite .........seriously i dont have a clue what ur are telling me

First the constant in the eq is $x^2$

Second how should i get rid of the other terms.....

4. Originally Posted by zorro
This is the equation that i have ended up with

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

I am using the Ratio test $\frac{u_{n+1} }{u_n} = k$ ....if $k<1 (convergent)$ $k>1(divergent)$ ....

$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\lim_{n \to \infty} \frac{x^2}{n+3 \sqrt{n+2}} . \frac{n+2 \sqrt{n+1}}{1}$

= $x^2 \frac{n+2 \sqrt{n+1}}{n+3 \sqrt{n+2}}$ .........I am stuck here what should i do next?????
hi zorro,
check your work again,
what is $u_{n}$?, $u_{n+1}$?

Spoiler:

$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\lim_{n \to \infty} \frac{x^2}{(n+2) (\sqrt{n+2})} \times \frac{(n+1) (\sqrt{n+1})}{1}$ $=\lim_{n \to \infty} \frac{x^2}{(n+2)^{3/2}} \times \frac{(n+1)^{3/2}}{1} = x^2$

5. and rewrite this series as...

$\sum_{n=0}^{\infty} \frac{x^{2n}}{(n+1)^{3/2}}$

6. ## Sorry i made a mistake in the question which i posted

Originally Posted by dedust
hi zorro,
check your work again,
what is $u_{n}$?, $u_{n+1}$?

Spoiler:

$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\lim_{n \to \infty} \frac{x^2}{(n+2) (\sqrt{n+2})} \times \frac{(n+1) (\sqrt{n+1})}{1}$ $=\lim_{n \to \infty} \frac{x^2}{(n+2)^{3/2}} \times \frac{(n+1)^{3/2}}{1} = x^2$

Sorry I posted the question falsely the question is this

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

7. It still doesn't matter, the answer is as I stated earlier $-1\le x\le 1$

8. Originally Posted by matheagle
It still doesn't matter, the answer is as I stated earlier $-1\le x\le 1$
But how would u do it ...........i am not able to get $x^2$

9. Originally Posted by zorro
Sorry I posted the question falsely the question is this

$\sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2) \sqrt{n+1}}$
$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\lim_{n \to \infty} \frac{x^2}{(n+3) (\sqrt{n+2})} \times \frac{(n+2) (\sqrt{n+1})}{1}$ $=\lim_{n \to \infty} \frac{x^2}{(n\sqrt{n+2}+3\sqrt{n+2})} \times \frac{(n\sqrt{n+1}+2\sqrt{n+1})}{1} =$ $\lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} =$ $\lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} \times$ $\frac{\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}}$ $= \lim_{n \to \infty} \frac{x^2}{(\sqrt{1+2n^{-1}}+3\sqrt{n^{-2}+2n^{-2}})} \times \frac{(\sqrt{1+n^{-1}}+2\sqrt{n^{-2} +n^{-3}})}{1} = x^2$

it will converge if . . .

10. $\frac{u_{n+1}}{u_n}= \frac{x^{2n+2}}{(n+3)\sqrt{n+3}}\frac{(n+2)\sqrt{n +1}}{x^{2n}}= \frac{x^{2n+2}}{x^{2n}}\frac{n+2}{n+3}\frac{\sqrt{ n+1}}{\sqrt{n+2}}$
$= x^2\frac{n+2}{n+3}\sqrt{\frac{n+1}{n+2}}$

Is that clearer?

Take the limit, as n goes to infinity, by dividing numerator and denominator of each fraction by n.

11. Originally Posted by dedust
$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\lim_{n \to \infty} \frac{x^2}{(n+3) (\sqrt{n+2})} \times \frac{(n+2) (\sqrt{n+1})}{1}$ $=\lim_{n \to \infty} \frac{x^2}{(n\sqrt{n+2}+3\sqrt{n+2})} \times \frac{(n\sqrt{n+1}+2\sqrt{n+1})}{1} =$ $\lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} =$ $\lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} \times$ $\frac{\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}}$ $= \lim_{n \to \infty} \frac{x^2}{(\sqrt{1+2n^{-1}}+3\sqrt{n^{-2}+2n^{-2}})} \times \frac{(\sqrt{1+n^{-1}}+2\sqrt{n^{-2} +n^{-3}})}{1} = x^2$

it will converge if . . .
How do u know it will converge as $x^2$ is neither >1 or <1 .........

12. Originally Posted by zorro

I am using the Ratio test $\lim_{n \to \infty}\frac{u_{n+1} }{u_n} = k$ ....if $k<1 (convergent)$ $k>1(divergent)$ ....

here, $\lim_{n \to \infty}\frac{u_{n+1} }{u_n} = x^2$

13. Originally Posted by dedust
here, $\lim_{n \to \infty}\frac{u_{n+1} }{u_n} = x^2$
since the test couldnt find if the series is divergent or not i am now going to use Raabe Ratio test

$\lim_{n \to \infty} n \left(\frac{u_n}{u_{n+1}}-1 \right)$ = $\lim_{n \to \infty} n \left(\frac{1}{x^2} -1 \right)$ =

$\lim_{n \to \infty} n \left(\frac{1 - x^2}{x^2} \right)$ = $\lim_{n \to \infty} \left(\frac{n -n. x^2}{n.x^2} \right)$

dividing by $1/n^2$ weget

= 0

$\lim_{n \to \infty} n \left(\frac{u_n}{u_{n+1}}-1 \right)$ = 0<1 $\therefore$ divergent