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Say I have a power series $\sum_{k=0}^\infty a_k x^k $ which converge uniformly on $\left[0, 1\right)$ . Now I need to prove that series $\sum_{k=0}^\infty a_k $ are convergent. My idea is to use equivalent Cauchy form $ \forall \epsilon\ \exists N \text{ such that }\sup_\left[0, 1\right) |S_n(x)-S_m(x)|<\epsilon\quad \forall m,n\ge N\ $ where $S_n = \sum_{k=0}^n a_k x^k$. Because of continuity of $P(x) = |S_n(x)-S_m(x)|$ we see that $ \sup_\left[0, 1\right) P(x) = \sup_\left[0, 1\right] P(x) $ and this way it can be proved that $\sum_{k=0}^\infty a_k $ convergent as Cauchy sequence.

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  • $\begingroup$ You have the correct idea. See this answer for a similar problem. $\endgroup$ – Winther Jan 9 '15 at 18:39
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Given $\epsilon>0$

$$|\sum_{k=n}^{m}a_k|=|\sum_{k=n}^{m} a_k-\sum_{k=n}^{m} a_kx^k+\sum_{k=n}^{m} a_kx^k|\leq |\sum_{k=n}^{m} a_k-\sum_{k=n}^{m} a_kx^k|+|\sum_{k=n}^{m} a_kx^k|$$

Take now $N$ large enough such that $|\sum_{k=n}^{m} a_kx^k|<\epsilon/2$ for all $x\in[0,1)$ and all $n,m>N$.Then move $x$ close enough to $1$ such that $|\sum_{k=n}^{m} a_k-\sum_{k=n}^{m} a_kx^k|<\epsilon/2$

Therefore $$|\sum_{k=n}^{m}a_k|<\epsilon$$ for $n,m>N$.

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