Continuity of a power-series function

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SUMMARY

The function g(x) = ∑(1/n^0.5)(x^(2n) - x^(2n+1)) is proven to be continuous on the interval [0, 1]. The series can be rewritten as g(x) = (1 - x) ∑(1/n^0.5)x^(2n), where the power series p(x) = ∑(1/n^0.5)x^(2n) has a radius of convergence of 1. The continuity of p(x) on the interval (-1, 1) is established, necessitating further examination of the convergence at the endpoint x = 1.

PREREQUISITES
  • Understanding of power series and their convergence properties
  • Familiarity with the concept of continuity in real analysis
  • Knowledge of the radius of convergence and related theorems
  • Basic proficiency in manipulating infinite series
NEXT STEPS
  • Research the properties of power series and their continuity
  • Study theorems related to the convergence of series at endpoints
  • Explore the implications of the Weierstrass M-test for uniform convergence
  • Investigate the behavior of series at boundary points, particularly at x = 1
USEFUL FOR

Mathematics students, educators, and researchers focusing on real analysis, particularly those studying continuity and convergence of power series.

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Homework Statement



Prove the function:
[itex]g(x)=\sum_{n=1}^{\infty }\frac{1}{^{n^{0.5}}}(x^{2n}-x^{2n+1})[/itex]

is continuous in [0,1].2. The attempt at a solution

I tried to look at this functions as:

[itex]g(x)=(1-x)\sum_{n=1}^{\infty }\frac{1}{^{n^{0.5}}}x^{2n}[/itex]

but I couldn't find a way solving it from here.
Finding the radius of convergence (which is 1) didn't help a lot...
 
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Yes, writing it as
$$g(x) = (1-x) \sum_{n=1}^\infty \frac{1}{n^{0.5}} x^{2n}$$
is a good start. Let us introduce the notation
$$p(x) = \sum_{n=1}^\infty \frac{1}{n^{0.5}} x^{2n}$$
This is a power series. If its radius of convergence is 1, do you know a theorem which tells you that p is continuous on (-1,1)? If so, then you only need to worry about the point x = 1. Does the series converge at x = 1?
 

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