Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$ [duplicate]

Solution 1:

Perhaps one of the most elementary ways to prove it: since $\,n\geq 1\,\,\forall n\in\mathbb{N}\,$ , we can put$$\sqrt[n]{n}=1+c_n\,,\,c_n\geq0\Longrightarrow n=(1+c_n)^n\geq \frac{n(n-1)}{2}c_n^2$$using the binomial expansion ,so that$$0<c_n\leq\sqrt\frac{2}{n-1}$$and now just apply the squeeze theorem and get $\,c_n\to 0\,$, which is precisely what we need.

Solution 2:

Claim: For each $a>1$, there exists $N$ such that $n<a^n$ for all $n>N$.

Proof: Write $a=1+b$. By the binomial theorem, $a^n=(1+b)^n\geq \frac{1}{2}n(n-1)b^2$ when $n\geq 2$. Thus $\frac{a^n}{n} \geq \frac{1}{2}(n-1)b^2$. It follows that if $N$ is at least $2/b^2+1$, then when $n>N$, $n<a^n$.

As a consequence, for each $a>1$, there exists $N$ such that $1\leq n^{1/n}<a$ for all $n>N$, and this implies that $\lim\limits_{n\to\infty}n^{1/n}=1$.

Solution 3:

$$\lim_{n\rightarrow\infty}\sqrt[n]n$$ $$=\lim_{n\rightarrow\infty}e^{\frac{\ln(n)}{n}}$$ and as we know that $\lim_{n\rightarrow\infty}\frac{\ln(n)}{n} = 0$ {apply l'Hospital's rule}
So $$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$

Solution 4:

One could use the fact that for a sequence of positive terms, if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} \root n\of {a_n}$ and the two limits are equal. A proof of this general fact can be found in these notes of Pete L. Clark. This result can also be found in many analysis texts; e.g., baby Rudin.

That your sequence has limit $1$ is easily shown using the above fact. A detailed proof that your sequence has limit $1$, based on the proof of the above fact, can be found in this thread.