# Sum of $\sum^\infty_ \mathrm{n=0} a_nx^n$

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 $$\frac{1}{1-x}$$

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 $$\dfrac{1}{1-x}$$.

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

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:

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