1. ## Convergence Problem

This is the equation that i have ended up with

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

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

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

= $\displaystyle x^2 \frac{n+2 \sqrt{n+1}}{n+3 \sqrt{n+2}}$ .........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 $\displaystyle x^2$
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 $\displaystyle (n+1)^{3/2}$
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 $\displaystyle x^2$
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 $\displaystyle (n+1)^{3/2}$
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 $\displaystyle x^2$

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

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

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

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

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

Spoiler:

$\displaystyle \lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\displaystyle \lim_{n \to \infty} \frac{x^2}{(n+2) (\sqrt{n+2})} \times \frac{(n+1) (\sqrt{n+1})}{1}$$\displaystyle =\lim_{n \to \infty} \frac{x^2}{(n+2)^{3/2}} \times \frac{(n+1)^{3/2}}{1} = x^2 5. and rewrite this series as... \displaystyle \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+1)^{3/2}} 6. ## Sorry i made a mistake in the question which i posted Originally Posted by dedust hi zorro, check your work again, what is \displaystyle u_{n}?, \displaystyle u_{n+1}? Spoiler: \displaystyle \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \displaystyle \lim_{n \to \infty} \frac{x^2}{(n+2) (\sqrt{n+2})} \times \frac{(n+1) (\sqrt{n+1})}{1}$$\displaystyle =\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

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

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

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

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

$\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2) \sqrt{n+1}}$
$\displaystyle \lim_{n \to \infty} \frac{u_{n+1}}{u_n}$ = $\displaystyle \lim_{n \to \infty} \frac{x^2}{(n+3) (\sqrt{n+2})} \times \frac{(n+2) (\sqrt{n+1})}{1}$$\displaystyle =\lim_{n \to \infty} \frac{x^2}{(n\sqrt{n+2}+3\sqrt{n+2})} \times \frac{(n\sqrt{n+1}+2\sqrt{n+1})}{1} =$$\displaystyle \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} = $$\displaystyle \lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} \times$$\displaystyle \frac{\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}}$$\displaystyle = \lim_{n \to \infty} \frac{x^2}{(\sqrt{1+2n^{-1}}+3\sqrt{n^{-2}+2n^{-2}})} \times \frac{(\sqrt{1+n^{-1}}+2\sqrt{n^{-2} +n^{-3}})}{1} = x^2 it will converge if . . . 10. \displaystyle \frac{u_{n+1}}{u_n}= \frac{x^{2n+2}}{(n+3)\sqrt{n+3}}\frac{(n+2)\sqrt{n +1}}{x^{2n}}= \frac{x^{2n+2}}{x^{2n}}\frac{n+2}{n+3}\frac{\sqrt{ n+1}}{\sqrt{n+2}} \displaystyle = x^2\frac{n+2}{n+3}\sqrt{\frac{n+1}{n+2}} Is that clearer? Take the limit, as n goes to infinity, by dividing numerator and denominator of each fraction by n. 11. Originally Posted by dedust \displaystyle \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \displaystyle \lim_{n \to \infty} \frac{x^2}{(n+3) (\sqrt{n+2})} \times \frac{(n+2) (\sqrt{n+1})}{1}$$\displaystyle =\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} =$$\displaystyle \lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} =$$\displaystyle \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$$\displaystyle \frac{\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}}$$\displaystyle = \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 $\displaystyle x^2$ is neither >1 or <1 .........

12. Originally Posted by zorro

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

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

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

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

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

dividing by $\displaystyle 1/n^2$ weget

= 0

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