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Suppose that for all integers $n ≥ 0$ and $0 ≤ a_n ≤ 1$.

Suppose also that $0 ≤ x < 1$.

Given these conditions, I'm trying to prove that if $\sum^\infty_ \mathrm{n=0} a_nx^n$ converges, its sum is not greater than $\frac{1}{1-x}$

Of course, the proof for convergence has been addressed already on this site. It does not discuss how the sum might be determined.

Based on how this is written, I would assume the answer is x-dependent (x cannot be outside of [0,1) ), but I have no idea how you reach $\frac{1}{1-x}$

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3 Answers 3

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You do not need to find a closed-form for the series in order to compare it with $\frac{1}{1-x}$. Actually, it is impossible to do so without knowing what the $a_n$'s are.

You simply work with the partial sum $$ \sum_{n=0}^Na_nx^n\le \sum_{n=0}^Nx^n\tag{1} $$ and then take the limit.

You can actually prove something stronger. (1) tells you that under the assumptions for $a_n$ and $x$, the series $\displaystyle \sum_{n=0}^\infty a_nx^n$ must converge and not greater than $\dfrac{1}{1-x}$.

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You get $$\left|\sum_{k=0}^\infty a_kx^k \right|\leq\sum_{k=0}^\infty |a_k||x|^k\leq\sum_{k=0}^\infty |x|^k=\sum_{k=0}^\infty x^k.$$

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If $0 \leq a_n \leq 1$, then multiplying by $x^n$, $0 \leq a_n x^n \leq x^n$.

Taking the series of those terms:

$$\sum_{n=0}^{\infty} a_nx^n \leq \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}.$$

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