Does Convergence at Zero Imply Global Convergence for Power Series?

  • Thread starter Thread starter Hummingbird25
  • Start date Start date
  • Tags Tags
    Convergence Series
Hummingbird25
Messages
84
Reaction score
0
Dear all

If a series e.g. a power series results in x convergering towards zero, can then one conclude that this series converge for all number if let's x belongs to R?

Sincerely Yours
Hummingbird25
 
Physics news on Phys.org
what do you mean by R? the ratio between terms?
 
I believe Hummingbird was referring to the set of real numbers (often denoted by R). The answer to the question (if I read it correctly, I had to read it a few times) is no also, look up "radius of convergence".
 
Hummingbird25 said:
Dear all

If a series e.g. a power series results in x convergering towards zero, can then one conclude that this series converge for all number if let's x belongs to R?

Sincerely Yours
Hummingbird25
This makes no sense at all. "A power series results in x converging towards zero"? First of all, x does not "converge" toward anything. It is a variable. Second, I don't know what you mean by saying "a power series results" in that.

If I really had to guess, I would guess you are asking about the "ratio test". If, for any series of positive numbers
\Sum_{n=0}^\infnty a_n the sequence \frac{a_{n+1}}{a_n} converges to any number less than 1, then the series converges.

From that it follows that if, for the power series \Sum_{n=0}^\infty a_nx^n[/itex] and some specific x, the ratio \left|\frac{a_{n+1}}{a_n}\right|\left|x\right| is less than 1 then the series converges for that x. In particular, if \left|\frac{a_{n+1}}{a_n}\right| converges to 0 then the above will converge to 0 &lt;1 for <b>all</b> x and so the power series converges for all x.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top