Does Convergence at Zero Imply Global Convergence for Power Series?

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Hummingbird25
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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
 
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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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