# If $a_n\ge nb_n$ and the sequence $(b_n)$ is unbounded, then the differences $a_{n+1}-a_n$ are also unbounded

Let $(a_n)_{n\geq 1}$ and $(b_n)_{n\geq1}$ be sequences of positive numbers such that $a_n\geq n b_n$ for all $n >1$.

Prove that if $(a_n)_{n\geq 1}$ is increasing and $(b_n)_{n\geq 1}$ is unbounded, then the sequence $(c_n)_{n\geq 1}$, given by

$$c_n=a_{n+1}-a_n$$ is also unbounded.

• Suppose $(c_n)$ were bounded. Then find a bound for $a_n$ (depending on $n$). From that, deduce that $(b_n)$ must be bounded. – Daniel Fischer Aug 17 '14 at 22:15
• how to find this bound for $a_n$? can you give the exact solution please. – Student Aug 17 '14 at 22:16

Suppose that $(c_n)_n$ is bounded and let $c\in \Bbb R$ such that $c_n\leq c$ then $a_{n+1}\leq a_n+c$, by recurrence we obtain $a_n\leq nc+a_0$, so for all $n\geq 1$ $a_n\leq c+\dfrac{a_0}{n}\leq c+a_0$, now we have $b_n\leq \dfrac{a_n}{n}\leq c=a_0$ $i.e$ $(b_n)_n$ is boundede.
Assuming that $\{c_n\}_{n\in\mathbb{N}}$ is bounded by $K$, we have: $$a_{n} = (a_n-a_{n-1})+\ldots+(a_{1}-a_0)+a_0 \leq Kn+a_0,$$ giving: $$b_{n} \leq K+\frac{a_0}{n},$$ contradiction.
Suppose $c_n \le L$. Then $a_{n+1} -a_n \le L$, of course, and hence $a_{n+k} -a_n \le kL$.
Then we have $a_n + kL \ge a_{n+k} \ge (n+k) b_{n+k}$. Dividing across by $k$ gives ${a_n \over k} + L \ge (1 + {n \over k}) b_{n+k}$.
Hence $L \ge \limsup_k b_{n+k}$.