If $\{a_n\}$ is bounded and non-decreasing, prove that $\liminf b_n = 0$, $b_n = n(a_{n+1} - a_{n})$

Question

Let $\{a_n\}$ be a bounded and non-decreasing sequence of reals, and $b_n = n(a_{n+1} - a_{n})$.

(a) Show that $\liminf b_n = 0$

(b) Give an example of a sequence $\{a_n\}$ such that $\{b_n\}$ diverges.

I know that $a_{n+1} \geq a_{n}, \forall n$,

Hence, $a_{n+1} - a_{n} \geq 0, \forall n$.

Also, we have that $\exists B >0$ s.t $|a_n| \leq B, \forall n$.

Hence, $b_n \geq 0$ and $|b_n| \leq n2B$

If $\{a_n\}$ is bounded, then $\{a_n\}$ converges by Bolzano Weierstrass Theorem, since $\{a_n\} \subseteq [-B,B]$

Now, since $\{a_n\}$ converges, there exists a subsequence $\{a_{n_k}\}$ of bounded variation, hence $\forall \epsilon >0, \exists N >0, \forall n_k\geq N$,

$|a_{n_{k+1}} - a_{n_k}| < \epsilon$

But I can't proof that the $\liminf b_n =0$.

Answer 1

Assume that $c = \liminf b_n > 0$. Then there is a $N \in \Bbb N$ such that $$ n(a_{n+1} - a_{n}) = b_n > \frac c2 > 0 \implies a_{n+1} - a_n > \frac{c}{2n} $$ for all $n \ge N$. Summing the last inequality gives $$ a_{n} = a_N + \sum_{j=N}^{n-1} (a_{j+1} - a_j) > a_N + \frac c2 \sum_{j=N}^{n-1} \frac 1j $$ for $n \ge N$. Since the harmonic series diverges, it follows that $a_n \to \infty$, contrary to the assumption that the sequence $(a_n)$ is bounded.