Show that $a_n^{1/n}$ converges as $n\to\infty$ Suppose that a sequence ${a_n : n = 1,2,\cdots}$ real numbers is such that
$a_n\geq 1$ for all $n ≥ 1$ and
$$
a_{n+m}\leq a_na_m\quad \rm{for~ all}~n\geq1, m\geq1.
$$
Show that $a_n^{1/n}$ converges as $n\to\infty$

My solution:
By taking log, we have

$$ \log{a_{n+m}}\leq \log{a_na_m}=\log{a_n}+\log{a_m}\quad \rm{for~ all}~n\geq1, m\geq1.
$$

So we have
\begin{align}
\log{a_2}&amp;\leq \log{a_1}+\log{a_1}=2\log{a_1}
\\
\log{a_3}&amp;\leq \log{a_1}+\log{a_2}\leq3\log{a_1}
\\
\cdots
\\
\log{a_n}&amp;\leq n\log{a_1}
\end{align}
So, 
$$
\log{a_n}^{1/n}=\frac{1}{n}\log{a_n}
$$
So I can prove $\log a_n$ is bounded but cannot to prove it's monotonic which can sufficiently lead to $\log a_n$ converges. How to deal with it?

Or if we cannot prove monotonic, how to prove the limit exists?
we have
$$a_{n+m}\leq a_na_m\quad $$
taking m=n,  we get
$$a_{2n}\leq a_n^2\quad $$

taking square roots on both sides, we get

$$ a_{2n}^{1/2}\leq a_n\quad$$

taking $n^{th}$ roots on both sides , we get

$$ a_{2n}^{1/2n}\leq a_n^{1/n}\quad$$

let

$$f(n)=a_n\quad$$

then , we have
$$f(2n)\leq f(n)\quad$$

which is a monotic decreasing sequence and hence converges
we have
$$a_{n+m}\leq a_na_m\quad $$
taking m=n,  we get
$$a_{2n}\leq a_n^2\quad $$

taking square roots on both sides, we get

$$ a_{2n}^{1/2}\leq a_n\quad$$

taking $n^{th}$ roots on both sides , we get

$$ a_{2n}^{1/2n}\leq a_n^{1/n}\quad$$

let

$$f(n)=a_n\quad$$

then , we have
$$f(2n)\leq f(n)\quad$$

which is a monotic decreasing sequence and hence converges