$\sum \frac{n^2}{3^nz^n}\\$

$\frac{1}{R}=\lim_{n->\infty}\sqrt[n]{|\sum \frac{n^2}{3^nz^n}|}\\$

what to do now?

2. Originally Posted by transgalactic
$\sum \frac{n^2}{3^nz^n}\\$

$\frac{1}{R}=\lim_{n->\infty}\sqrt[n]{|\sum \frac{n^2}{3^nz^n}|}\\$

what to do now?
I don't know why you have the summation inside the square root. Correct application of the nth root test means that the given series will converge for values of z that satisfy $\lim_{n \rightarrow +\infty} \left| \frac{n^{2/n}}{3z}\right| < 1 \Rightarrow \frac{1}{3|z|} \lim_{n \rightarrow +\infty} n^{2/n} < 1$.

Then you have to test each endpoint.

3. i cant understand the numenator
we have $n^{\frac{n}{2}}$
so we get "infinty by power 0"

what is the result of this expression

i think its undefined form

?

4. Originally Posted by transgalactic
i cant understand the numenator
we have $n^{\frac{n}{2}}$
so we get "infinty by power 0"

what is the result of this expression

i think its undefined form

?

No, you don't have n^(n/2) but n^(2/n) and the limit of this thing is 1, so the lmit you're looking for is (1/3)(1/|z|) = 1/R and etc.

Tonio

5. sorry i ment that
n^(2/n)

i cant understand why it equals 1
we have (infinity)^0
why its 1?

6. Originally Posted by transgalactic
sorry i ment that
n^(2/n)

i cant understand why it equals 1
we have (infinity)^0
why its 1?

Since power series and stuff belongs at least to calculus II , I assume you've already had calculus I: limits, continuity and stuff...right?
Then you must know that n^(1/n) --> 1 when n --> oo, but then
n^(2/n) = [n^(1/n)]^2 --> 1*1 = 1 by arithmetic of limits.

Tonio

7. ok now i have
$\frac{1}{3|z|}$

i dont know how to deside if it converge or diverge

so what if its less then 1

by what law we decided that it converged?

8. Originally Posted by transgalactic
ok now i have
$\frac{1}{3|z|}$

i dont know how to deside if it converge or diverge

so what if its less then 1

by what law we decided that it converged?

Hmmm...rechecking your question I realize that what you posted is NOT a power series in z since z appears in the denominator, so this is at most a Laurent series, a very different stuff belonging to complex analysis.
What did you really mean to ask here? Is z a variable or just a parameter?

Tonio

9. z is a complex number
here it acts as a parameter

?

10. Originally Posted by transgalactic
z is a complex number
here it acts as a parameter

?

Oh, I see: so z is a constant. Well, since it MUST be (1/3)(1/|z|) < 1, from here we see that the series converges whenever |z| > 1/3.
Now just check the end points 1/3 and -1/3 (hint: divergence at both)

Tonio

11. why it converges
whenever |z| > 1/3

??
by what law?

i know that a series converges if the limit is a constant
it diverges if its infinity

12. Originally Posted by transgalactic
why it converges
whenever |z| > 1/3

??
by what law?
Inequalities...??! If 1/(3|z|) < 1 then multiplying by |z| you get |z| > 1/3...

Tonio

13. ok i understand
but the formula is
1/R=1/(3|z|)
R=3|z|
so if going stricktly by the formula i get undefined expression 3|z|

that it uniformly coverges from -1/3<z<1/3

but i get a different answer by the formula

14. Originally Posted by transgalactic
ok i understand
but the formula is
1/R=1/(3|z|)
R=3|z|
so if going stricktly by the formula i get undefined expression 3|z|

that it uniformly coverges from -1/3<z<1/3

but i get a different answer by the formula

True, because as I pointed out before what you ngave is not a power series! What you gave is a palin series with a parameter (constant) z, as you said. There's no convergence radius here, then.
For something to be a proper power series the powers of the VARIABLE have to be positive integers.

Tonio

15. Originally Posted by transgalactic
ok i understand
but the formula is
1/R=1/(3|z|)
R=3|z|
so if going stricktly by the formula i get undefined expression 3|z|