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Hello this problem I have no idea what can I do.

Let $\sum a_n x^n$ be a power series with finite convergence radius $r$. Prove that if $\sum a_n r^n$ converges, then $\sum a_n x^n$ converges uniformly in $[0,r]$.

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This is the Abel theorem for real numbers! Here is the link of that theorem: – Tigran Hakobyan Oct 31 '11 at 19:35
Dear August, Accepting answers is a nice thing to do, but you shouldn't forget to upvote (clicking the upwards arrow to the left of questions/answers) stuff you found helpful! – J. M. Nov 1 '11 at 16:56
up vote 2 down vote accepted

From the convergence of $\sum a_n x^n$ in $x = r$ it follows that $a_n r^n \to 0$. So in particular, the sequence is bounded, say by $M > 0$, that is $|a_n r^n| \leqslant M$ or $|a_n| \leqslant M r^{-n}$. Now let $l \in [0, r)$, then for each $k$ we have

$$|a_k l^k| \leqslant |a_k| |l|^k \leqslant \left (\frac{|x|}{r} \right )^k$$

By assumption we have $|z|/r < 1$. Therefore by the comparison test the series converges absolutely, hence converges.

Now, if a series converges uniformly Cauchy, then it converges uniformly. Now the above series converges uniformly if for any $\varepsilon > 0$, there is a positive integer $N$ such that if $N_0 \leqslant j \leqslant m$ then

$$\left |\sum_{k = j}^m a_n x^n \right | < \varepsilon$$

for all $x \in [0, r)$. Now

$$\left |\sum_{k = j}^m a_n x^n \right | \leq \sum_{k = j}^m |a_n| |x|^n.$$

This can be made easily smaller than any $\varepsilon > 0$. Do you see how?

So, now we should extend the result to $x = r$. As said in the comments, Abel's theorem allows to extend the convergence to the boundary if $r = 1$.

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If you replace "for all $x\in [0,r)$" by "for all $x\in [0,r]$, what changes? – Weltschmerz Nov 1 '11 at 16:58
@Weltschmerz Then we would have $|a_k l^k| \leqslant 1$ and then we wouldn't have convergence. – Jonas Teuwen Nov 1 '11 at 22:45

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