How to show that $\lim_{n\rightarrow\infty}(n^{2}+1)^{\frac{1}{n}} = 1$? Prove $\lim_{n\rightarrow\infty}(n^{2}+1)^{\frac{1}{n}} = 1$ . I am wondering if I can use the ratio test to prove this? When I use the ratio test I get the following:
$\frac{((n+1)^{2}+1)^{\frac{1}{n}}}{(n^{2}+1)^\frac{1}{n}}$
But this is painfully difficult to handle, am I doing the right things?
Or should I do the following:
Let $\epsilon>0$. Then let us consider the sequence, ${\frac{n^{2}+1}{(1+\epsilon)^{n}}}$. Then we have by the ratio test that:
$\frac{a_{n+1}}{a_{n}} = \frac{(n^{2}+2)}{(1+\epsilon)^{n+1}}\frac{(1+\epsilon)^{n}}{n^{2}+1} = \frac{n^{2}+2}{n^{2}+1}\frac{1}{1+\epsilon}$. And since the term on the left goes to 1, as n goes to infinity, we would have that this ratio test would just equal $\frac{1}{1+\epsilon}$, which is less than 1, and hence this sequence converges to 0.
So from this knowledge we can say that ${\frac{n^{2}+1}{(1+\epsilon)^{n}}} < 1$, and so, ${{n^{2}+1} < {(1+\epsilon)^{n}}}$. And so if we raise LHS and RHS to $\frac{1}{n}$ we have proved it?
 A: $$
\lim_{n\to\infty}(n^2+1)^\frac{1}{n}=\lim_{n\to\infty}e^{\frac{\ln(1+n^2)}{n}}=\lim_{n\to\infty}e^{\frac{\ln(n^2)}{n}}=\lim_{n\to\infty}e^{\frac{2\ln n}{n}}=e^0=1
$$
A: Note that
$$n^{2/n} \leqslant (n^2+1)^{1/n} \leqslant (2n^2)^{1/n} = 2^{1/n} n^{2/n}$$
By the squeeze  theorem it follows that $\lim_{n \to \infty} (n^2+1)^{1/n}  = 1$, since $2^{1/n} \to 1$ and $n^{2/n} \to 1$ as $n \to \infty$.

That $2^{1/n} \to 1$ is well-known and easy to prove. To show that $n^{2/n}\to 1$, we have $n^{2/n} \geqslant 1$ for all $n \in \mathbb{N}$ and for $n > 2$ using the binomial theorem we have
$$n^2 = (1+ (n^{2/n}-1))^n> \frac{n(n-1)(n-2)}{6}(n^{2/n} -1)^3$$
Thus, for all $n > 2$,
$$0 < n^{2/n}- 1 < \sqrt[3]{\frac{6n}{(n-1)(n-2)}} \underset{n \to \infty}\longrightarrow 0$$
A: We know for a seuence $\{x_n\}_n^\infty$ of positive reals, if $\lim \frac{x_{n+1}}{x_n}=L \in [0,\infty]$ then $\lim {x_{n}}^{\frac1 n}=L$.
So, $$\lim \frac{(n+1)^2+1}{n^2+1}=\lim \frac{(1+\frac1 n)^2+\frac 1 {n^2}}{1+\frac1 {n^2}}=1\\ \implies \lim (n^2+1)^{\frac 1 n}=1$$
