# Power series question (ROC)

• Apr 28th 2011, 01:12 PM
sedam7
Power series question (ROC)
hello,

I see now that i did these power series by pattern and now I'm stuck big time :D

Problem is ...

$\displaystyle \sum_{n=1}^{\infty} \frac {n}{3^nx^{2n}}$

so i found R as

$\displaystyle R = \lim_{n\to \infty} \big{|} \frac {a_n}{a_{n+1}}\big{|} = ... = 3$

and than i sow x is at power -2n and here is my problem...

should I do it like this : (to see region of convergence)

$\displaystyle |x|< 3$

or like this ...

$\displaystyle |x^{-2}|< 3$

$\displaystyle |x|< \frac{1}{\sqrt{3}}$

or I am way of with this ?

can someone please explain to me what to do when i have "x" like i do here :D

• Apr 28th 2011, 01:31 PM
topsquark
Quote:

Originally Posted by sedam7
hello,

I see now that i did these power series by pattern and now I'm stuck big time :D

Problem is ...

$\displaystyle \sum_{n=1}^{\infty} \frac {n}{3^nx^{2n}}$

so i found R as

$\displaystyle R = \lim_{n\to \infty} \big{|} \frac {a_n}{a_{n+1}}\big{|} = ... = 3$

and than i sow x is at power -2n and here is my problem...

should I do it like this : (to see region of convergence)

$\displaystyle |x|< 3$

or like this ...

$\displaystyle |x^{-2}|< 3$

$\displaystyle |x|< \frac{1}{\sqrt{3}}$

or I am way of with this ?

can someone please explain to me what to do when i have "x" like i do here :D

My first question to you (and I don't know the answer) is if you can use that test when the problem is not a power series.

In any event you have your logic backward. $|x| < \frac{1}{\sqrt{3}}$ is not the radius of convergence. The series converges on $|x| > \frac{1}{\sqrt{3}}$. So this is not a "radius" of convergence at all.

-Dan
• Apr 28th 2011, 01:51 PM
sedam7
hm... what do you mean ? can I use that test when series is not power series ? (I used D'Alambert criteria for testing of convergence :D )

$\displaystyle \sum_{n = 0}^\infty \frac{n}{3^n} x^{-2n}$

so i got

$a_n = \frac {n}{3^n}$

Than ...

$R=\lim_{x \to \infty} |\frac{a_n}{ a_{n+1}} | = \lim_{x \to \infty} \frac{\frac{n}{3^n}}{\frac{n+1}{3^n \cdot 3}} = 3$

then since in power series |x| <R is region of convergence .... i assumed that |x|<3 but... if I put x = 4 ... than .... using D'Alambert test i get ... .

$q= \lim_{x \to \infty} \frac{a_{n+1}}{a_n} = .... = \lim_{x \to \infty} \frac{n+1}{64n} = \frac {1}{64} < 1$

so series converge ? or I probably should go to sleep now :D and try it tomorow :D
• Apr 28th 2011, 02:03 PM
topsquark
Quote:

Originally Posted by sedam7
hm... what do you mean ? can I use that test when series is not power series ? (I used D'Alambert criteria for testing of convergence :D )

$\displaystyle \sum_{n = 0}^\infty \frac{n}{3^n} x^{-2n}$

so i got

$a_n = \frac {n}{3^n}$

Than ...

R=\lim_{x \to \infty} |\frac{a_n}{ a_{n+1}} | = \lim_{x \to \infty} \frac{\frac{n}{3^n}}{\frac{n+1}{3^n \cdot 3}} = 3

then since in power series |x| <R is region of convergence .... i assumed that |x|<3 but... if I put x = 4 ... than .... using D'Alambert test i get ... .

q= \lim_{x \to \infty} \frac{a_{n+1}}{a_n} = .... = \lim_{x \to \infty} \frac{n+1}{64n} = \frac {1}{64} < 1

so series converge ? or I probably should go to sleep now :D and try it tomorow :D

Well as I said, I didn't know if the test were more general than power series.

Look, if nothing else convinces you, try calculating x = 0.5. This is inside what you call your radius of convergence. It does not converge. Whereas, as you say, x = 4 does converge.

-Dan
• Apr 28th 2011, 02:11 PM
sedam7
Thanks :D I was to tired and there's language issue ... and... Thank you very very much :D

you have saved me :D (Bow)(Bow)(Bow)(Bow)(Bow)(Bow)(Bow)

$|\frac{1}{x}|<3$

$x\in (-\infty, -3] U [3, +\infty)$ :D

I'm so stupid... sorry that i bother everybody with such stupidity ....
• Apr 28th 2011, 02:25 PM
topsquark
Quote:

Originally Posted by sedam7
Thanks :D I was to tired and there's language issue ... and... Thank you very very much :D

you have saved me :D (Bow)(Bow)(Bow)(Bow)(Bow)(Bow)(Bow)

$|\frac{1}{x}|<3$

$x\in (-\infty, -3] U [3, +\infty)$ :D

I'm so stupid... sorry that i bother everybody with such stupidity ....

You aren't stupid. You made a mistake. Stupidity would have been not asking for help.

-Dan
• Apr 28th 2011, 02:39 PM
Plato
Quote:

Originally Posted by sedam7
$|\frac{1}{x}|<3$
$x\in (-\infty, -3] U [3, +\infty)$ :D

You need to rethink that answer.
The series converges for $x=0.6$ the sum is $\frac{674}{4}$.
• Apr 28th 2011, 02:54 PM
Plato
Quote:

Originally Posted by topsquark
This was bugging me and since I put in the work I thought I'd share. So the series converges for
$|x| > \frac{1}{\sqrt{3}}$

That is correct. I will point out that the absolute value of the root test works nicely.
• Apr 28th 2011, 03:45 PM
yeKciM
yes, converge for $|x| > \frac{1}{\sqrt{3}}$

and points $\displaystyle \frac {1}{\sqrt{3}}$ are not included :D
• Apr 29th 2011, 01:37 AM
sedam7
okay i see what i have done :( sorry :D
now I'm confused... in this problem i have to find radius of convergence ? what to say on this now ? heheheheh

because these series converge in zero and for $|x| > 1/ \sqrt{3}$ ?
• Apr 29th 2011, 10:59 AM
HallsofIvy
The difficulty is, as TopQuark said initially, that your series is NOT a power series. It does not have x to positive powers and so does NOT have a "radius of convergence". With x to a negative power, you get, as shown, that it converges outside some radius. Now, it is possible that the series is actually Sum (n/3^n) x^n?
• Apr 29th 2011, 03:00 PM
sedam7
hmmmm....
i was doing some problems from exams last year :D and i come to this one :D

Quote:

Find radius of convergence of the series:

$\displaystyle \sum_{n=1} ^{\infty} \frac{n}{3^n x^{2n}}$

then examine the convergence of the boundary points of intervals.
so i have done what I wrote there and didn't know what to do else so i posted it here if someone could explain it ... actually as I look at it better ... three years ago was more or less the same problem ...

Quote:

Find radius of convergence of the series:

$\displaystyle \sum_{n=1} ^{\infty} \frac{n}{4^n x^{2n}}$

then examine the convergence of the boundary points of intervals.
actually it's same just with $4^n$ ....

now I'm very very confused :D soon I'll have exam and can't find any of them to see what to do with this :D or I'll just hope something like this I will not get :D
• Apr 29th 2011, 03:18 PM
Plato
Quote:

Originally Posted by sedam7
hmmmm....
i was doing some problems from exams last year :D and i come to this one :D so i have done what I wrote there and didn't know what to do else so i posted it here if someone could explain it ... actually as I look at it better ... three years ago was more or less the same problem ...
actually it's same just with $4^n$ ....
now I'm very very confused :D soon I'll have exam and can't find any of them to see what to do with this :D or I'll just hope something like this I will not get :D

No one can blame you for being confused.
I personally think that the instructor who set this question is badly confused.

That ‘old’ series converges if $|x|>\dfrac{1}{2}$ or in the set $\left( { - \infty , - \frac{1}{2}} \right) \cup \left( {\frac{1}{2},\infty } \right)$.
That is unbounded set. What is the radius of an unbounded set?
You see why that instructor’s question is nonsense?