Function represented by power series

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SUMMARY

The discussion centers on the continuity of functions represented by power series, specifically the series ∑(k=0 to ∞) a_k(x-a)^k, which converges within a radius R. It is established that the function f is continuous on the interval (a-R, a+R) due to the uniform continuity of power series within their radius of convergence. A critical point raised is the invalid assumption regarding the interchange of limits in the proof, highlighting the necessity of careful application of limit properties in analysis.

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Homework Statement


If a function f is represented by the power series ∑(k=0 to ∞) a_k(x-a)^k, with a radius of convergence )<R<∞, then f is continuous on the interval (a-R, a+R)


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The Attempt at a Solution



I don't know if my proof is loose at some point or not, because it seems so easy. Is there anything that is obviously not right about my proof? Thanks very much

So choose x=m in (a-R, a+R), the series converges on the interval (a-R,a+R)
then we would have
lim(x->m) ∑(k=0 to ∞) a_k(x-a)^k = ∑(k=0 to ∞) (a_k)lim(x->m) (x-a)^k = ∑(k=0 to ∞) (a_k)(m-a)^k = f(m).

So f is continuous on any arbitrary m on (a-R, a+R) therefore ∑(k=0 to ∞) a_k(x-a)^k is continuous on (a-R, a+R)
 
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It is possible to have series of continuous functions that converges to a non-continuous function. To prove the result, you must use the fact that a power series is uniformly continuous within its radius of convergence.
 
Your proof is invalid, because you cannot assume that

[tex]\lim_{x \to m} \lim_{n \to \infty} \sum_{k=0}^n a_k (x-a)^k = \lim_{n \to \infty} \lim_{x \to m} \sum_{k=0}^n a_k (x-a)^k[/tex].

Consider [itex]x_{m,n}[/itex] = 0 when n < m and 1 when n >= m. Then [itex]\lim_{n \to \infty} \lim_{m \to \infty} x_{m,n} \neq \lim_{m \to \infty} \lim_{n \to \infty} x_{m,n}[/itex].
 
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