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I have some problems with this, because this sequence could converge to some point, or goes to infinity, only this two possibilities, and really I don´t know how to prove that no other possibility can happen, that´s my problem. The sequence is such that exist a fixed positive constant C, such that for every m,n integers $ a_m + a_n < a_{m + n} + C $

prove that $ (a_n)/n $ converge to a point, or goes to infinity $

and some idea to prove in general that other of two possibilities can´t happen?

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    What do you mean by converging to a point?2011-08-12
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    I mean that given any epsilon, exist an N such that n>N implies that $$ \left| {a_n - L} \right| < \varepsilon $$ where L is the point2011-08-12
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    Are all the $a_n$ nonnegative?2011-08-12
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    I don´t know, could be that goes to -infinity, but consider also that case, i.e (in the case of to infinity given any A>0 exist an N , such that if n>N then $ a_n > A $ and similar to -infinity, only that 3 cases, could happen2011-08-12
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    You probably need to demonstrate $a_n/n$ is monotone and bounded from below, then use the least upper bound axiom (flipped) and monotone convergence theorem.2011-08-11
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    Somehow I submitted my comment backwards through time. Is math.se hooked up to a relativistic CTC by any chance?2011-08-11
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    This is called **Fekete subadditive lemma**, see http://en.wikipedia.org/wiki/Subadditivity#Properties2011-08-12

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Replacing $a_n$ by $a_n - C$ affects neither to the convergence nor to the limiting value of $a_{n} / n$. In this case, we have $$a_{m} + a_{n} \leq a_{m+n},$$ and this gives a simple lower bound for $a_{n} / n$, namely $a_1$. Now this shows that $s = \limsup_{n\to\infty} a_{n} / n$ is either $+\infty$ or a finite constant.

Note that we can find a subsequence $n_k$ such that $a_{n_k} / n_{k} \to s$. Now fix $k$ and write $n = q n_k + r$, where $q$ and $r$ are integers depending on both $n$ and $k$ such that $1 \leq r \leq n_k$. Then $$ a_{n} \geq a_{qn_{k}} + a_{r} \geq q a_{n_{k}} + a_{r}.$$ Dividing both sides by $n$ gives $$ \frac{a_{n}}{n} \geq \frac{q a_{n_{k}} + a_{r}}{qn_{k} + r} = \frac{a_{n_{k}} + (a_{r} / q)}{n_{k} + (r/q)}$$ for $n$ large so that $q > 0$. Then taking $\liminf_{n\to\infty}$ to both sides yields $$ \liminf_{n\to\infty} \frac{a_{n}}{n} \geq \frac{a_{n_{k}}}{n_{k}}.$$ But since this is true for any $k$, taking $k \to \infty$ proves that $a_{n} / n$ tends to either $+\infty$ or a finite value as $n \to \infty$.

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Try the below method:change variables $b_n=C-a_n$,we get $b_{n+m}n}\{ \frac{b_k}{k}\}$,$b$ exsits (may be infinity).Similiarly,we have $b'=\lim_{n\longrightarrow \infty}\sup_{k>n}\{ \frac{b_k}{k}\}$,and $b\leq b'$.Next we show that $b=b'$.By definition of $b$,$ \forall \epsilon >0,\exists N,s.t.|\frac{b_N}{N}-b|<\epsilon$.By definition of $b'$,there is a subsequnce $\{\frac{b_{n_r}}{n_r}\}$ which converges to $b'$.Note $\forall n_r,\exists k_r,q_r\in \mathbb{N}(0\leq q_r

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    Oops, I think I'm late...2011-08-12
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    @sos440:nothing serious:)2011-08-12