Prove $\sum_{n=0}^{\infty} \lvert x(1-x)^n \rvert $ $\leq$ $\sum_{n=0}^{\infty} (1/2)^n $ $x \in [0,1] $. Does this hold for all partial sums?

Question

This feels true but I don't know how to prove it. However, I have concerns when $x$ is close to $1$, since then the summation on the left will grow faster than the function on the right.

Will there be an $x$ and $n$ at which the left side will be greater than the right? If so, is there any sequence of positive numbers (not dependent on $x$) that converges that I can replace $\left(\frac{1}{2}\right)^n$ with so this inequality holds true for any $x$ and $n$?

Bruh this editing took forever. Is there any way to view what my equations look like before posting?

Answer 1

$$\sum_{n=0}^{+\infty}| x(1-x)^n |=\sum_{n=0}^{+\infty} x(1-x)^n = 1$$

for all $x \in ]0,1]$

$$\sum_{n=0}^{+\infty} (1/2)^n =2 >\sum_{n=0}^{+\infty} x(1-x)^n=1 $$

as for $x=0$ , $ 0<2 $

Chose an integer $n$:

$$\sum_{k=0}^{n}| x(1-x)^k |=1-(1-x)^{n+1} $$

$$\sum_{k=0}^{n} (1/2)^k=2(1-(\frac{1}{2})^{n+1}) $$

and so $$\sum_{k=0}^{n} (1/2)^k -\sum_{k=0}^{n}| x(1-x)^k |= 1-(\frac{1}{2})^{n } +(1-x)^{n+1}>0$$