How prove $x_n=(1+a_{n})^\frac1n $ is convergent if $1+a_{m+n}\leq (1+a_m)(1+a_n)$? assume  {$a_n$}$_{n=1}^\infty$ ,$a_n$ is none negative and real sequence that satisfied :$$1+a_{m+n}\leq (1+a_{m})(1+a_{n}) ,\quad m,n\in\mathbb N$$  how prove $x_n=(1+a_{n})^\frac1n $  is convergent? thanks in advance 
 A: Set $y_n:=1+a_n$ and note $y_n\geq 1$ by your assumptions, with
$$
y_{m+n}\leq y_my_n.
$$
The key trick is to use Euclidean division.
Fix $m\geq 1$.
Now for all $n$, do the Euclidean division 
$$
n=mq+r\qquad\mbox{with}\; 0\leq r<q.
$$
Then it is easily seen that
$$
y_n=y_{mq+r}\leq y_m^qy_r\leq y_m^q C
$$
with $C=\max\{y_r\;:\;0\leq r<q\}$.
Next 
$$
y_n^{1/n}\leq y_m^{q/n}C^{1/n}\leq y_m^{1/m}C^{1/n}.
$$
This holds for all $n\geq 1$, remember that $m$ is fixed.
Taking $\limsup$, we find:
$$
\limsup y_n^{1/n}\leq y_m^{1/m}.
$$
This holds for all $m\geq 1$, so now we can take $\inf$ and $\liminf$:
$$
\limsup y_n^{1/n}\leq \inf y_m^{1/m} \leq \liminf y_m^{1/m}.
$$
Finally, we see that $\liminf y_n^{1/n}=\liminf y_n^{1/n}$, so the sequence converges and moreover:
$$
\lim y_n^{1/n}=\inf y_n.
$$
Note: This is how you prove that the formula $\lim \|T^n\|^{1/n}$ makes sense for a bounded linear operator. Then you prove it is equal to the spectral radius.
A: Set $b_n= 1+ a_n$
Take a $N$ and let's we will see how much the term $b_N$ has grown in terms of a fixed $n$
For that reason divide $N$ with $n$
$$N=nm+u$$
Therefore $b_N=b_{nm+u} \leq b_{nm}b_u \leq b_n^m b_u$ 
So we get $b_N^{\frac{1}{N}} \leq b_n^{\frac{m}{nm+u}}b_u^{\frac{1}{N}}\leq b_n^{\frac{m}{nm+u}}(\max _{u \in \{0,1,2 \cdots, n)}b_u)^{\frac{1}{N}}$
By taking $$\lim \sup _N$$ at both sides we have
$$\lim \sup b_N^{\frac{1}{N}} \leq b_n^{\frac{1}{n}}$$ Take now 
$\lim \inf $ in respect of $n$ and you are done.
ADDED: This result sometimes is seen in the sub additive form $a_{m+n}\leq a_m+a_n$
and it goes by the name Fekete's lemma, which not surprisingly (just found out by googling) is discussed in math.stackexchange before.
