# Prove inequality: $\frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2})^2}+\cdot\cdot\cdot+\frac{1}{(\sqrt{2})^n}\leq\frac{1}{\sqrt{2}-1}, \forall n\in\mathbb{N}$

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!

• Do you know how to sum a geometric series? Apr 19, 2020 at 14:56
• @user619894 thank you so much, i've just looked it up! Apr 19, 2020 at 15:03
• What you're facing is the need to strengthen the induction hypothesis. Although in this case it is best to recognize it as a geometric series, you should learn to try this technique. For example, if you try to prove $1/3+1/3^2+1/3^3+\cdots+1/3^n<1/2$ by induction, using the desired claim as the hypothesis would lead nowhere, but you can get it to work with the stronger hypothesis $1/3+1/3^2+1/3^3+\cdots+1/3^n≤1/2·(1-1/3^n)$. Try it! Finding such strengthenings is an art, but can be extremely useful when you cannot find closed-forms. Apr 26, 2020 at 9:18

From high-school, you have $$x+x^2+\dots+x^n=x(1+x+\dots+x^{n-1})=x\,\frac{1-x^n}{1-x}.$$