# how prove $x_n=(1+a_{n})^\frac1n$ is convergent? such that {$a_n$} satisfied in following conditions

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

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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.

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Good answer, but the question already uses $x_n$ so it would be clearer to use a different variable name. – Thomas Andrews Feb 1 '13 at 19:51
Thanks, Thomas. Edited. – 1015 Feb 1 '13 at 20:04

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.

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Minor detail: your $u$ depends on $N$ but can take only finitely many values since $0\leq u< n$. So actually you need to bound $b_u$ by the max of the $b_u$'s involved, before you can take the $\limsup$. – 1015 Feb 1 '13 at 19:19
Thanks, I tottaly understimated the remainder of the division @julien – clark Feb 1 '13 at 19:21