How would i prove the following inequality:
$\frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2})^2}+\cdot\cdot\cdot+\frac{1}{(\sqrt{2})^n}\leq\frac{1}{\sqrt{2}-1}, \forall n\in\mathbb{N}$.
Maybe the inequality is obvious but I'm missing it. Anyway, here's my attempt for proof by induction:
$\textbf{Base case}: n=1\\ \frac{1}{\sqrt{2}} \leq \frac{1}{\sqrt{2}-1}\\ \textbf{Induction step}: n\rightarrow n+1\\ \frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2})^2}+\cdot\cdot\cdot+\frac{1}{(\sqrt{2})^n}+\frac{1}{(\sqrt{2})^{n+1}}\leq \frac{1}{\sqrt{2}-1} $
And now I'm lost, I don't know what to do when there are no variables on the right side of the inequality. Feel free to post some other proof that you can come up with!
Hint:
From high-school, you have $$x+x^2+\dots+x^n=x(1+x+\dots+x^{n-1})=x\,\frac{1-x^n}{1-x}.$$