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Show that if the sequence ${a_n}$ is bounded then the power series $\sum a_nx^n$ converges absolutely for $|x|<1$.

I haven't the slightest idea how to prove this. Does anyone have any thoughts on this?

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Suppose the $a_n$ are uniformly bounded in absolute value by $M$. Then

$$\sum |a_n x^n| = \sum |a_n| |x|^n \le \sum M|x|^n = \frac{M}{1-x}$$

when $|x|<1$.

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Hint: for all $N \geq 0$ $$ 0 \leq \sum_{n=0}^N \lvert a_n x^n\rvert = \sum_{n=0}^N \lvert a_n\rvert \lvert x\rvert^n \leq \sum_{n=0}^N \lVert a\rVert_\infty \lvert x\rvert^n = \lVert a\rVert_\infty\sum_{n=0}^N \lvert x\rvert^n $$ and conclude using the radius of convergence of the series $\sum \lvert x\rvert^n$.

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