Let $(a_n)$ be a positive and monotonically increasing sequence that
satisfies$$ a_{n \cdot m} \geq n a_m \; \; \; \; \text{for} \; \; n,m \in
\mathbb{N} $$Suppose $\sup_{n \in \mathbb{N}} \left\{ \dfrac{a_n}{n} \right\} < +
\infty $, Prove ${\bf carefully}$ that $\left( \dfrac{ a_n }{n}
\right) $ ${\bf converges}$
Attempt:
By hypothesis, if $\alpha$ is the supremum of the sequence, then $\dfrac{a_n}{n} \leq \alpha $ so we observe that $(a_n/n)$ is bounded. If we can prove that is monotonic, then we are done. Let $b_n = a_n/n$. Then, we have
$$ \dfrac{b_{n+1} }{b_n} = \dfrac{n}{n+1} \cdot \frac{a_{n+1}}{a_n} > \dfrac{n}{n+1} $$
Which leads nowhere. But, if we use the property of the sequence with $m=1$ we see that
$$ a_n \geq n a_1 $$
And in particular with $n$ replaced by $n+1$ one sees that $a_{n+1} \geq (n+1) a_1 $ or that $\dfrac{ a_{n+1} }{n+1} \geq a_1 $. Now, Id be tempted to say that
$$ \dfrac{ a_{n+1} }{n+1} – \dfrac{ a_n }{n} \geq a_1 – a_1 = 0 $$
but unfortunaly the inequality is not always true. Am I on the right directio nto solve this problem?Any hint/suggestion is welcome!
Best Answer
Clearly $\limsup_{k \to \infty} \frac{a_k}{k} \le \sup_n \frac{a_n}{n}$. Fix $n \ge 1$. For $k \ge 1$, write $k = nq+r$ for $0 \le r \le n-1$. Then $\frac{a_k}{k} = \frac{a_{nq+r}}{nq+r} \ge \frac{a_{nq}}{nq+r} \ge \frac{qa_n}{qn+r}$. So, $\liminf_{k \to \infty} \frac{a_k}{k} \ge \frac{a_n}{n}$. It follows that $\liminf_{k \to \infty} \frac{a_k}{k} \ge \sup_n \frac{a_n}{n}$.