please use the the limit of difinition to prove the following:

Question

$$\lim_{n\to \infty}\sqrt[n]{n^2+1}=1$$ I try to use the binomial theorem: let $$\sqrt[n]{n^2+1}=1+y_n\\$$ then $$n^2+1=(1+y_n)^n$$ then $$n^2+1>1+ny_n$$ get $$y_n<n$$ based on the definition of limit,I can get $$|\sqrt[n]{n^2+1}|<|1+y_n-1|<|y_n|<|n|<\varepsilon$$ let $$N=[\varepsilon]$$ then right here I find It doesn't work. I don't konw where has gone wrong,or the method is not applicable to this question. I ask for your help.

Answer 1

For $n\geq 4$ we have $$n^2+1=(1+y_n)^n=1+ny_n+\frac{n(n-1)}{2!}y_n^2+\frac{n(n-1)(n-2)}{3!}y_n^3+...$$$$\geq \frac{n(n-1)(n-2)}{3!}y_n^3\geq \frac{(n-2)^3}{6}y_n^3$$

Now, $$y^3_n\leq 6\frac{n^2+1}{(n-2)^3}\leq\frac{12n^2}{(n-2)^3}\leq\frac{12n^2}{\big(n/2\big)^3}\leq \frac{96}{n}.$$

Hence, for given any $\epsilon>0$ we have, $$0\leq y_n\leq \frac{\sqrt[3]{96}}{\sqrt[3]{n}}<\epsilon\text{ for all }n\geq \bigg\lfloor\frac{96}{\epsilon^3}\bigg\rfloor+1$$ So, $\lim y_n=0$.