Proving $\displaystyle{\lim_{n \to\infty}\sqrt[n]{n} = 1}$ by Contradiction and Using the Binomial Expansion

Question

Given that $$\displaystyle{\lim_{n \to\infty}\sqrt[n]{n} = L}$$ exists such that L ≥ 1. I need to prove by contradiction that L = 1. Suppose for the sake of contradiction that L > 1 and $\delta$ = L - 1 > 0. Using the binomial expansion, I have showed that $$\displaystyle{L^n = (1 + \delta)^n ≥ ((n(n-1))/2) \delta^2}$$ Given $$\displaystyle{((n(n-1))/2)≥(n^2)/4}$$ for every $n ≥ 2$, if we take $n ≥ \max (2, 8/\delta^2)$ we obtain that $L^n ≥ 2n$. Now, I am required to obtain a contradiction from the conditions above and show that $L = 1$. I am really confused on how to proceed with obtaining a contradiction, any helpful hint would be greatly appreciated.

Answer 1

Let $\sqrt[n]{n}=1+y_n$, then $$n = (1+y_n)^n \geq 1+ \frac{n(n-1)}{2}y_n^2.$$ It follows that $$|\sqrt[n]{n}-1| = |y_n| < \sqrt{\frac{2}{n}},$$ which means $\lim_{n\to\infty} \sqrt[n]{n}=1.$