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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    $\begingroup$ 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. $\endgroup$ – anon Aug 11 '11 at 22:07
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    $\begingroup$ Somehow I submitted my comment backwards through time. Is math.se hooked up to a relativistic CTC by any chance? $\endgroup$ – anon Aug 11 '11 at 22:14
  • $\begingroup$ What do you mean by converging to a point? $\endgroup$ – gary Aug 12 '11 at 2:52
  • $\begingroup$ 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 point $\endgroup$ – Daniel Aug 12 '11 at 3:07
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    $\begingroup$ Are all the $a_n$ nonnegative? $\endgroup$ – Dylan Moreland Aug 12 '11 at 3:09

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 $n \to \infty$.


Try the below method:change variables $b_n=C-a_n$,we get $b_{n+m}<b_n+b_m$.Besides,it is easily seen that $\{ \frac{a_n}{n}\}$ and $\{\frac{b_n}{n}\}$ have the same convergent behavoir(both are convergent to some point,or both go to infinity,or both are divergent).The convergence of $\{\frac{b_n}{n}\}$ is a well known result.You can do it on your own or see as follows:let $b=\lim_{n\longrightarrow \infty}\inf_{k>n}\{ \frac{b_k}{k}\}$,$b$ 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 $\forall n_r,\exists k_r,q_r\in \mathbb{N}(0\leq q_r <N),s.t.n_r=k_r N+q_r$.Beside $b_{n_r}<k_rb_N+b_{q_r}$.Note $\{b_{q_r}\},\{q_r\}$ are bounded,thus taking limits with respect to $r$ in $\frac{b_{n_r}}{n_r}<\frac{k_r b_N+b_{q_r}}{k_r N+q_r}$,we get $b'\leq \frac{b_N}{N}<b+\epsilon$.Since $\epsilon $ is arbitary,we get $b'\leq b$.As a result,$b=b'$,which exactly means that the sequence $\{\frac{b_n}{n}\}$ converges.As it it unknown whether $b$ is finite or infinite,we get the two cases you have just pointed out.

  • $\begingroup$ Oops, I think I'm late... $\endgroup$ – Sangchul Lee Aug 12 '11 at 5:28
  • $\begingroup$ @sos440:nothing serious:) $\endgroup$ – user14242 Aug 12 '11 at 5:36

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