The inequality $\lim \frac{1}{2} - \frac{x}{4} + ... + \frac{(-1)^{n} x^{n}}{2^{n+1}} \le \left( \frac{1}{7-7x^{3}} \right)^{1/3}$

Question

I made a problem:

Prove that if $|x| < 1$ and $n$ is even, then $$ \lim_{n \rightarrow \infty} \frac{1}{2} - \frac{x}{4} + \frac{x^{2}}{8} - \frac{x^{3}}{16} + \frac{x^{4}}{32} - \frac{x^{5}}{64} + ... + \frac{(-1)^{n} x^{n}}{2^{n+1}} $$ $$ \le \left( \frac{1}{1-x^{3}} \right)^{1/3} \left( \frac{1/8}{1-1/8} \right)^{1/3} = \left( \frac{1}{7-7x^{3}} \right)^{1/3} $$

I would like to know if this problem can be solved in a shorter way than using Holder's?

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We can use Holders inequality by considering 3 infinite sequences:

$$1, x, x^{2}, x^{3}, ... $$ $$1,-1,1,-1,1,-1,1,... $$ $$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, ... $$

And by $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = (1/3) + (1/3) + (1/3) = 1$. After that we will see two geometric series which converges to $1/(1-x^{3})$ and $1/7$ respectively, then the proof is done.

Answer 1

Counterexample:

$n:=2 ~ , ~~\displaystyle x:=-\frac{1}{2} :$

$\displaystyle \frac{1}{2}-\frac{x}{4}+\frac{x^2}{8} = 0.65625$

$\displaystyle \left(\frac{1}{7-7x^3}\right)^{1/3} = 0.5026316274194358675807...$

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$n\to\infty$ :

$7-7x^3 \leq (x+2)^3$

Counterexample with $~\displaystyle x:=-\frac{1}{2}~$:

$7-7x^3 = 7.875$

$(x+2)^3 = 3.375$