# this sequence converges?

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 $and some idea to prove in general that other of two possibilities can´t happen? • You probably need to demonstrate$a_n/n$is monotone and bounded from below, then use the least upper bound axiom (flipped) and monotone convergence theorem. – anon Aug 11 '11 at 22:07 • Somehow I submitted my comment backwards through time. Is math.se hooked up to a relativistic CTC by any chance? – anon Aug 11 '11 at 22:14 • What do you mean by converging to a point? – gary Aug 12 '11 at 2:52 • 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 – Daniel Aug 12 '11 at 3:07 • Are all the$a_n$nonnegative? – Dylan Moreland Aug 12 '11 at 3:09 ## 2 Answers 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.

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