Is it true that every bounded sequence with the following property converges?

Question

Is it true that every bounded sequence $\{a_n\}$ of real numbers such that $|{a_n - a_{n-1}}|<1/n$ for all $\ge2$ is convergent?

Answer 1

Let's try this again: take the half-harmonic sequence $1/2,1/4,1/6,...$. Every tail end of this sequence has infinite sum, since the whole sequence diverges. So I can define a sequence as follows: $a_n=\sum_{i=1}^n (-1)^{f(n)}1/(2i)$ where $f$ is determined recursively as follows: $f(1)=1$ and $$f(n)=\left\{\begin{matrix} f(n-1),\text{ if } -1<a_{n-1}+(-1)^{f(n-1)}1/(2n)<1\\ -f(n-1),\text{ otherwise}\end{matrix}\right.$$ So an initial segment of our sequence is $1/2,1/2+1/4,1/2+1/4+1/6,1/2+1/4+1/6-1/8,1/2+1/4+1/6-1/8-1/10,...$. The point is that the sequence fluctuates between $-1$ and $1$ forever, using divergences of the half-harmonic sequence. So this is not convergent, and is bounded, but the successive differences have absolute value $1/(2n)$.

Answer 2

No. Because the harmonic series is divergent, you can create a sequence that just bounces from end to end of an interval - just keep summing terms $\frac{1}n$ until you get to some fixed upper bound, then subtracting terms $\frac{1}n$ until you reach the lower bound. So, you end up with lengthening oscillations of constant amplitude, so the series cannot converge. For instance, the partial sums of the following sequence satisfy this, but do not converge:

$$\begin{align}&+\frac{1}1\\ &-\frac{1}2-\frac{1}3 \\&+\frac{1}4+\frac{1}5+\frac{1}6+\frac{1}7\\&-\frac{1}8-\frac{1}9-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ldots\end{align}$$

where the sign flops every power of two. In each row, the partial sums change by somewhere between $\frac{2^i}{2^i}=1$ and $\frac{2^{i}}{2^i+1}=\frac{1}2$, so the sequence can't be Cauchy and thus does not converge. The sequence must be bounded, since, if we took the absolute value of sum of any pair of consecutive rows $$\sum_{k=2^i}^{2^{i+1}-1}\frac{1}{k}-\frac{1}{2k}-\frac{1}{2k+1}=\sum_{k=2^i}^{2^{i+1}}\frac{1}{4k^2+2k}$$ which, of course, converges, being the sum of reciprocals of squares. Thus, the series is bounded and this is a counterexample to your claim.

Answer 3

Yet one more example:

$a_n = \sin \big[ \ln( n+1 ) \big]. $

Because for every $x, y \in \mathbb{R}$ we have

$|\sin x - \sin y| \leqslant |x - y|,$

we get

$| a_n - a_{n-1} | \leqslant | \ln(n+1) - \ln n | = \ln \left( 1 + \frac{1}{n} \right) < \frac{1}{n}.$

It is not difficult to prove however, that $a_n$ is bounded and divergent.

This example takes slightly more analysis knowledge, but is less formally problematic.

Answer 4

Set

$a_1 = 0$

$a_2 = 1/3$

Now $a_3 = 1/3 + 1/4$

Now keep adding $1/5$, $1/6$ and so on till it crosses $1$.

Now keep subtracting till it crosses $0$, then flip again to cross $1$ and keep going back and forth.

This sequence satisfies $|a_n - a_{n-1}| \lt 1/n$, and has a subsequence which converges to $0$ and a subsequence which converges to $1$.

EDIT: Oops same as Meelo's answer.