# Thread: Test the convergence of the series

1. ## Test the convergence of the series

Question :

Test the convergence of the series

$\displaystyle \frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...$ for all x

2. Originally Posted by zorro
Question :

Test the convergence of the series

$\displaystyle \frac{x^2}{3 \sqrt{2}} + \frac{x^4}{4 \sqrt{3}} + \frac{x^6}{5 \sqrt{4}} + ...$ for all x
The general term is $\displaystyle \frac{x^{2n}}{(n + 2) \sqrt{n + 1}}$. Now choose an appropriate test from the list of tests you have been taught and apply it.

3. ## I am stuck here

Originally Posted by mr fantastic
The general term is $\displaystyle \frac{x^{2n}}{(n + 2) \sqrt{n + 1}}$. Now choose an appropriate test from the list of tests you have been taught and apply it.

I am using the ratio test to test the series ...

$\displaystyle \lim_{n \to \infty} \frac{a_n + 1 }{a_n} = \lim_{n \to \infty} \frac{x^{2(n+1)}}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$

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

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

= $\displaystyle \lim_{n \to \infty} \frac{x^2 \left[(n + 2) \sqrt{n + 1} \right]}{(n + 3) \sqrt{n + 2}}$

I am stuck here!!!!!!

4. try to use root test.

5. ## I am using the root test

Originally Posted by Krizalid
try to use root test.

I am using the root test and have got this answer where i am stuck right now

6. Originally Posted by zorro
I am using the ratio test to test the series ...

$\displaystyle \lim_{n \to \infty} \frac{a_n + 1 }{a_n} = \lim_{n \to \infty} \frac{x^{2(n+1)}}{(n + 3) \sqrt{n + 2}} . \frac{(n + 2) \sqrt{n + 1}}{x^{2n}}$

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

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

= $\displaystyle \lim_{n \to \infty} \frac{x^2 \left[(n + 2) \sqrt{n + 1} \right]}{(n + 3) \sqrt{n + 2}}$

I am stuck here!!!!!!
From your last step, you can pull out the $\displaystyle x^2$ (because it has no $\displaystyle n$ in it), so

$\displaystyle \lim_{n \to \infty} \frac{x^2(n + 2) \sqrt{n + 1}}{(n + 3) \sqrt{n + 2}} = x^2\lim_{n\to\infty}\frac{\sqrt{(n+2)^2(n+1)}}{\sq rt{(n+3)^2(n+2)}}=x^2\lim_{n\to\infty}\frac{\sqrt{ n^3+5n^2+8n+4}}{\sqrt{n^3+8n^2+21n+18}}$

Spoiler:
Now pull the $\displaystyle n^3$ out of the square root to get:

$\displaystyle x^2\lim_{n\to\infty}\frac{n^{3/2}\sqrt{1+5/n+8/n^2+4/n^3}}{n^{3/2}\sqrt{1+8/n+21/n^2+18/n^3}}$

Now you can see that taking $\displaystyle n\to\infty$ gives you a limit of $\displaystyle 1$, (i.e. the whole thing equals $\displaystyle x^2$), so the series converges when $\displaystyle |x^2|<1$, so when $\displaystyle |x|<1$. You also need to check for convergence at the endpoints ($\displaystyle 1$ and $\displaystyle -1$).
Spoiler:
For $\displaystyle -1$, use the alternating series test. For $\displaystyle +1$, compare it to $\displaystyle \sum\frac{1}{n^{3/2}}$.
Spoiler:
Just kidding; two nested spoilers is enough.

7. ## Alternating series for -1

Originally Posted by redsoxfan325
From your last step, you can pull out the $\displaystyle x^2$ (because it has no $\displaystyle n$ in it), so

$\displaystyle \lim_{n \to \infty} \frac{x^2(n + 2) \sqrt{n + 1}}{(n + 3) \sqrt{n + 2}} = x^2\lim_{n\to\infty}\frac{\sqrt{(n+2)^2(n+1)}}{\sq rt{(n+3)^2(n+2)}}=x^2\lim_{n\to\infty}\frac{\sqrt{ n^3+5n^2+8n+4}}{\sqrt{n^3+8n^2+21n+18}}$

Spoiler:
Now pull the $\displaystyle n^3$ out of the square root to get:

$\displaystyle x^2\lim_{n\to\infty}\frac{n^{3/2}\sqrt{1+5/n+8/n^2+4/n^3}}{n^{3/2}\sqrt{1+8/n+21/n^2+18/n^3}}$

Now you can see that taking $\displaystyle n\to\infty$ gives you a limit of $\displaystyle 1$, (i.e. the whole thing equals $\displaystyle x^2$), so the series converges when $\displaystyle |x^2|<1$, so when $\displaystyle |x|<1$. You also need to check for convergence at the endpoints ($\displaystyle 1$ and $\displaystyle -1$).
Spoiler:
For $\displaystyle -1$, use the alternating series test. For $\displaystyle +1$, compare it to $\displaystyle \sum\frac{1}{n^{3/2}}$.
Spoiler:
Just kidding; two nested spoilers is enough.

Thanks for the reply....i am stuck again ....I know it is a silly question but could u please tell me on what should i use the alternate series test for

I know i need to use for -1 but which equation should i use

8. Originally Posted by zorro
Thanks for the reply....i am stuck again ....I know it is a silly question but could u please tell me on what should i use the alternate series test for

I know i need to use for -1 but which equation should i use
Since your series involves even powers of x, x = 1 and x = -1 give the same series. The alternating series test is not relevant because the terms do not alternate when x = -1. The comparison test, implied in the spoiler of an earlier reply, is all that is required.

9. ## My question wasnt that

Originally Posted by mr fantastic
Since your series involves even powers of x, x = 1 and x = -1 give the same series. The alternating series test is not relevant because the terms do not alternate when x = -1. The comparison test, implied in the spoiler of an earlier reply, is all that is required.
Mr fantastic i wanted to know where should i put in the comparison test is it to be put in the eq $\displaystyle \frac{x^2n}{n + 2 \sqrt{n + 1}}$ or in the result $\displaystyle x^2$

10. Originally Posted by zorro
Mr fantastic i wanted to know where should i put in the comparison test is it to be put in the eq $\displaystyle \frac{x^2n}{n + 2 \sqrt{n + 1}}$ or in the result $\displaystyle x^2$
Substitute x = 1 into the original series. Then use the comparison test on the resulting series, as shown in the spoiler of an earlier reply. This will also establish what happens when x = -1 (because substituting x = -1 gives the same series).

11. ## Is this correct?

Originally Posted by mr fantastic
Substitute x = 1 into the original series. Then use the comparison test on the resulting series, as shown in the spoiler of an earlier reply. This will also establish what happens when x = -1 (because substituting x = -1 gives the same series).

Substitute x = 1 in the original series we get

$\displaystyle \frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}$

$\displaystyle u_n = \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}}$ and $\displaystyle v_n = \frac{1}{n^{\frac{3}{2}}}$

$\displaystyle \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{\frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} }{\frac{1}{n^{\frac{3}{2}}}}$ = $\displaystyle \lim_{n \to \infty} \frac{(-1)^2n}{(n + 2) \sqrt{n + 1}} . n^{\frac{3}{2}}$

Is this correct????????

12. Sorry, my bad. I didn't notice it was all even powers. In that case, -1 and 1 are the same.

13. ## Is the answer correct ?

Originally Posted by redsoxfan325
Sorry, my bad. I didn't notice it was all even powers. In that case, -1 and 1 are the same.
In my last post i have done some step could u please check and let me know if i have done it correctly or no .....and what else do i need to do after this

14. Originally Posted by zorro
Substitute x = 1 in the original series we get

$\displaystyle \frac{(-1)^2}{3 \sqrt{2}} + \frac{(-1)^4}{4 \sqrt{3}} + \frac{(-1)^6}{5 \sqrt{4}}$

$\displaystyle u_n = \frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}}$ and $\displaystyle v_n = \frac{1}{n^{\frac{3}{2}}}$

$\displaystyle \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{\frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}} }{\frac{1}{n^{\frac{3}{2}}}}$ = $\displaystyle \lim_{n \to \infty} \frac{(-1)^{2n}}{(n + 2) \sqrt{n + 1}} . n^{\frac{3}{2}}$

Is this correct????????
So far, yes. Now calculate the limit. You should find that it equals 1, which tells you...

15. ## I am stuck inthe limit

Originally Posted by redsoxfan325
So far, yes. Now calculate the limit. You should find that it equals 1, which tells you...

Could u please tell me how could i get rid of $\displaystyle (-1)^2n$

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