Convergent Sequence and Cauchy Criterion- Counter Example

Question

Consider the sequence $\left \{ x_{n} \right \}$ that satisfies the condition: $$\left | x_{n+1}-x_{n} \right |< \frac{1}{2^{n}} \ \ \ for\ all\ n=1,2,3,...$$ Part (1): Prove that the sequence $\left \{ x_{n} \right \}$ is convergent.

Part (2): Does the result in part (1) hold if we only assume that $\left | x_{n+1}-x_{n} \right |< \frac{1}{n} \ \ \ for\ all\ n=1,2,3,...$?

For part (1), I proved that the sequence is Cauchy and hence it is convergent. For part (2), I feel like the sequence is not necessarily convergent. I am trying to come up with a sequence that is divergent, but satisfies the condition given in part (2). Any ideas?

Answer 1

For part (2) you can also use $x_n = \ln(n)$.

Note that by the MVT we have for some $c_n \in (n,n+1)$:

$$ \frac{\ln(n+1)-\ln(n)}{n+1-n}= \frac{1}{c_n} < \frac{1}{n} \,.$$

Answer 2

Let $x_1=1$, say, and let $x_{n+1}=x_n+\frac1{2n}$ for each $n\in\Bbb Z^+$.