How can I solve the limit: $\lim_{n\to\infty} \frac{((-1)^n+5)^{1/n}}{3}$ How can I solve the limit ($n \in \mathbb{N}$): $$\lim_{n\to\infty} \frac{((-1)^n+5)^{1/n}}{3}$$Is it ok to say that it is greater or equal to $\frac{4^{1/n}}{3}$ and lower or equal to $\frac{6^{1/n}}{3}$ and thus the limit $=\frac{1}{3}$ ?

Note:
Here's how to show that
$a^{1/n}
\to 1
$
for $a > 1$.
Since
$a > 1$,
$a^{1/n} > 1$.
Therefore
$a^{1/n} = 1+b$
with
$b > 0$.
Raising to the
$n$-th power,
$a = (1+b)^n
$.
By Bernoulli's inequality
(easily proved by induction),
$(1+b)^n
\ge 1+nb
$.
Since
$b = a^{1/n}-1
$,
we have
$\begin{array}\\
a
&=(1+b)^n\\
&\ge 1+nb\\
&=1+n(a^{1/n}-1)\\
\text{so}\\
a^{1/n}
&\le \dfrac{a-1}{n}+1\\
\text{or}\\
a^{1/n}-1
&\le \dfrac{a-1}{n}\\
&\to 0 \text{ as } n \to \infty\\
\end{array}
$
