Prove that the nth root converges to 1 with Binominal Theorem Prove with the use of the Binominal Theorem that: > **$$\sqrt{x}=(\sqrt[n] {\sqrt{x}})^x=(\sqrt[x] {\sqrt{x}}-1)+1)^x≥1+x(\sqrt[n] {\sqrt{x}}-1) \quad \forall x \in \mathbb{N} $$** > > This is the first step to prove that the sequence $\sqrt[n] {x}\to1 $. I have gotten so far: By using the Binominal Theorem the inequality can be written as: $$((\sqrt[n] {\sqrt{x}}-1)+1)^x= \sum_{l=0}^x \binom xl(\sqrt[n] {\sqrt{x}}-1)^{l}≥\binom x2(\sqrt[n] {\sqrt{x}}-1)^2$$ If n≥2 then this inequality holds true. The last part of the inequality can be written as: $$\binom x2(\sqrt[n] {\sqrt{x}}-1)^2 = \frac {x(x-1)}2(\sqrt[n] {\sqrt{x}}-1)^2 $$ Therefore: $\sqrt{x}≥ \frac {x(x-1)}2(\sqrt[n] {\sqrt{x}}-1)^2. $ This is where I get stuck, what would be the next logical step from here?

It's a little simpler than you are making it....

$$ \begin{align} ((\sqrt[x] {\sqrt{x}}-1)+1)^x &= \sum_{\ell=0}^x \binom x\ell(\sqrt[x] {\sqrt{x}}-1)^{\ell} \\\ &\ge \binom x0(\sqrt[x] {\sqrt{x}}-1)^0 + \binom x1(\sqrt[x] {\sqrt{x}}-1)^1 \\\ &=1+ x(\sqrt[x] {\sqrt{x}}-1) \end{align} $$