proofverification of $\lim_{n\rightarrow\infty}\sqrt[n]{n+s}=1$ where $s\ge 0$ and request for an easier proof

Question

I have basically copied the original proof that $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$ and I hope that there are no mistakes:

$x_n=\sqrt[n]{n+s}-1\geq0$

$n+s=(x_n+1)^n\geq1+\binom{n}{2}x_n^2=1+\frac{n(n-1)}{2}x_n^2$

Therefore $x_n\leq\sqrt{\frac{2(n+s-1)}{n(n-1)}}$

And therefore $0\leq\sqrt[n]{n+s}-1=x_n\leq \sqrt{\frac{2(n+s-1)}{n(n-1)}}$

From now on I have stoppped to copy the original proof

$\lim$ stands for $\lim_{n\rightarrow \infty}$

And because $\lim\frac{2(n-1)}{2(n-1)+2s}=\lim 1-\frac{2s}{2(n-1)+2s}=1$ And the rule if $\lim {a_n}=a>0$ then $\lim \sqrt[k]{a_n}=\sqrt[k]{a}$

I can say (I am not sure if I am really allowed to do that, some verification would be appreciated)

$$\lim 0\leq \lim\sqrt[n]{n+s}-1=\lim x_n\leq \lim \sqrt{\frac{2(n+s-1)}{n(n-1)}}$$

$$\lim 0\leq \lim\sqrt[n]{n+s}-1=\lim x_n\leq \lim \sqrt{\frac{2(n+s-1)}{n(n-1)}}\cdot 1$$

This is the step where I am asking for verification

$$\lim 0\leq \lim\sqrt[n]{n+s}-1=\lim x_n\leq \lim \sqrt{\frac{2(n-1)+2s}{n(n-1)}}\cdot \sqrt{\frac{2(n-1)}{2(n-1)+2s}}$$

$$\lim 0\leq \lim\sqrt[n]{n+s}-1=\lim x_n\leq \lim \sqrt{\frac{2}{n}}=0$$

Is the proof alright? I hope there is an easier one

Answer 1

Easier proof: $$\lim_{n\to\infty}\log(\sqrt[n]{n+s}) = \lim_{n\to\infty}\frac{\log(n+s)}n = 0.$$

Answer 2

I think an easier one, goes like this:$$\lim_{n\to \infty}\sqrt[n]{an}{=\lim_{n\to \infty}\sqrt[n]{a}\sqrt[n]{n}\\=\lim_{n\to \infty}1\cdot\sqrt[n]{n}\\=\lim_{n\to \infty}\sqrt[n]{n}}$$for $a>0$. Also$$\lim_{n\to \infty}\sqrt[n]{n}=\lim_{u\to \infty} e^{ue^{-u}}=e^0=1$$therefore$$\lim_{n\to \infty}\sqrt[n]{an}=1$$ and hence$$1=\lim_{n\to \infty}\sqrt[n]{n}\le\lim_{n\to \infty}\sqrt[n]{n+s}\le \lim_{n\to \infty}\sqrt[n]{2n}=1$$which implies that$$\lim_{n\to \infty}\sqrt[n]{n+s}=1$$