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

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 $$\liminf b_n =0$$.

• Hint: Assume that $\liminf b_n > 0$. – Martin R Jan 11 at 14:21
• ... Then there exists $\epsilon>0$ and $N\in\mathbb{N}$ such that $b_n\ge \epsilon$ for all $n\ge N$ ... – Song Jan 11 at 14:22

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.
• We will have that the series of consecutive terms will be bigger than the harmonic, and hence that series will diverge, but if it should converge since the series of consecutive terms will be convergent. But the problem is, that only a subsequence of $a_n$ can be of bounded variation if $a_n$ converges... how can I know that the whole sequence is of bounded var ? – Richard Clare Jan 11 at 14:45
• @RichardClare: Summing the above relationship gives $a_{n+1} > a_N + \frac c2 \sum_{j=N}^n \frac 1j \to \infty$ for $n \to \infty$, contrary to the assumption that $(a_n)$ is bounded. – Martin R Jan 11 at 14:50
• @RichardClare: Note also that generally, the divergence of the “tail” $(a_n)_{n \ge N}$ implies the divergence of the “entire” sequence $(a_n)$. – Martin R Jan 11 at 14:53
• Its not $a_{N+1}$ ? and why you didn't apply the sum both sides ? – Richard Clare Jan 11 at 14:54
• I got this: $\sum_{n = N}^{\infty}(a_{n+1} - a_n) > \frac c2 \sum_{j=N}^n \frac 1j \to \infty$ – Richard Clare Jan 11 at 14:57