1. ## interval of convergence

What is the interval of convergence?

2. Here.

3. Originally Posted by Nitz456
What is the interval of convergence?

we proceed by the ratio test ( i hope you know it, if not, say so)

let An = (n^2 * x^2n)/2^2n

then by the ratio test, we want

lim{n-->infinity} |A(n+1)/An| < 1

lim{n-->infinity} |A(n+1)/An|
= lim{n-->infinity} |[(n+1)^2 * x^(2n + 2)]/2^(2n + 2) * (2^2n)/(n^2 * x^2n)|
= lim{n-->infinity} |(2^2n)/2^(2n + 2) * [(n+1)^2]/n^2 * x^(2n + 2)/x^2n|
= lim{n-->infinity} |2^-2 * (n^2 + 2n + 1)/n^2 * x|
= (1/4)|x|

for (1/4)|x| < 1
we must have |x| < 4

so -4 < x < 4

so our interval of convergence is (-4,4)

now we check the end points to see if that should be a closed interval

plug in -4 and then 4 for x and see if we have convergence

Note: E is summation

E (n^2 * (4)^2n)/2^2n = E (n^2 * 2^4n)/2^2n = E n^2 * 2^2n

this diverges as n--> infinity

a similar thing happens when x = -4

so the interval of convergence is (-4,4)

hmm, i got a different answer from TPH, wonder where i went wrong

oh, i see! i put x^(2n + 2)/x^2n = x (see the blue writing above), but it was supposed to be x^2. ok.

so just follow TPH's way, it's shorter--and correct

4. Originally Posted by Jhevon
we proceed by the ratio test ( i hope you know it, if not, say so)
If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".

And then consider the case when x=0 (but that case is trivial).

hmm, i got a different answer from TPH, wonder where i went wrong
I think you forgot a square somewhere.
You get that,
|x|<4
But it should be,
|x^2|<4

5. Originally Posted by ThePerfectHacker
If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".

And then consider the case when x=0 (but that case is trivial).

I think you forgot a square somewhere.
You get that,
|x|<4
But it should be,
|x^2|<4
yeah, i already caught that, thanks

6. Originally Posted by ThePerfectHacker
If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".

And then consider the case when x=0 (but that case is trivial).
and thanks for this, i forgot that little detail.

and this is the day before my exam! oh boy!

7. Thanks kindly folks! I used the ratio test answer because we've not yet learned of the root test.

8. Originally Posted by Nitz456
Thanks kindly folks! I used the ratio test answer because we've not yet learned of the root test.
make sure you correct my error though, i have not corrected it

9. mos def, sir.