# Limit of Root: Shifted

It is a well known fact that: $$\lim_{N\to\infty}\sqrt[n]{n}=1$$ But what about the shifted one up: $$\lim_{N\to\infty}\sqrt[n+1]{n}=1$$ And what about the shifted one down: $$\lim_{N\to\infty}\sqrt[n]{n+1}=1$$

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Well, you can use the fact that $\sqrt[n]{n}>\sqrt[n+1]{n}>1$, in order to conclude that $\lim\limits_{n\to\infty}\sqrt[n+1]{n}=1$

As to the other question you added:

You can use the fact that $\forall p\exists q>p:\sqrt[p]{p}>\sqrt[q]{q+1}>1$, and conclude that $\lim\limits_{n\to\infty}\sqrt[n]{n+1}=1$

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Cool thx and what about shifted one down (see edited question) – Alexander Frei May 1 '14 at 20:16
Mhh but you cannot deduce the last since: $\lim_{n\to\infty} a_n b_n,\lim_{n\to\infty} a_n\text{ exist}\nRightarrow \lim_{n\to\infty}b_n\text{ exists}$ – Alexander Frei May 1 '14 at 21:09
@Freeze_S: I agree, but I did the exact opposite in this case. If to take your notation, then I did $\lim\limits_{n\to\infty}a_n$ exists and $\lim\limits_{n\to\infty}b_n$ exists $\implies\lim\limits_{n\to\infty}a_nb_n$ exists. – barak manos May 1 '14 at 21:13
Yes so I added your answer according to this if u don't mind – Alexander Frei May 1 '14 at 21:18
@Freeze_S: I do mind, actually, because what you wrote has got nothing to do with my answer. I might as well remove it all together, because I don't see how $\sqrt[n]{n}\sqrt[n]{n}$ is helpful here. – barak manos May 1 '14 at 21:23

Consider $f(n) = \sqrt[n+1]{n} = n^{1/(n + 1)}$. Then

$$\ln f(n) = \frac{\ln n}{n + 1} \to 0$$

as $n \to \infty$, so $f(n) \to 1$.

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One can use the same proof. But there is no need to go back to basics, since $$n^{1/(n+1)}=(n+1)^{1/(n+1)} \left(\frac{n}{n+1}\right)^{1/(n+1)},$$ and it is clear that $$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{1/(n+1)}=1.$$

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