does this sequence necessarily converge?

Question

Let $\{x_n\}$ be a real sequence that satisfies $|x_{n+1} - x_n| < \frac{1}{n}$ for all $n \geq 1$.

Suppose we know that $\{x_n\}$ is bounded, then must $\{x_n\}$ converge?

Answer 1

No. For example, starting at $x_1 = 0$ let $x_{n+1} = x_n + 1/n$ until $x_n > 1$, then $x_{n+1} = x_n - 1/n$ until $x_n < 0$, then $+$ again ... Because $\sum_n 1/n$ diverges, you will have infinitely many terms $> 1$ and infinitely many $< 0$.

Answer 2

No. Here is an example: $$x_n = \sin ( \sum_{k=1}^{n-1} \frac{1}{k} )$$

Use the equality $\sin a - \sin b = 2 \sin(\frac{a-b}{2}) \cos \frac{a+b}{2}$ and the inequality $|\sin \delta|\le |\delta|$ to see that $|x_{n+1}- x_n| < \frac{1}{n}$.

Answer 3

I had trouble understanding the sequences presented above, so I tried to come up with a simpler one.

Consider the sequence

$$x_n \ = \ \sum_{k = 1}^{n} \ \dfrac{1}{2k - 1}.$$

Then $(x_n) \ = \ \bigg(1, 1 + \dfrac{1}{3}, 1 + \dfrac{1}{3} + \dfrac{1}{5}, 1 + \dfrac{1}{3} + \dfrac{1}{5} + \dfrac{1}{7}, ...\bigg)$. Note that

$$\lvert x_{n + 1} - x_n \rvert \ = \ \bigg\lvert \sum_{k = 1}^{n + 1} \ \dfrac{1}{2k - 1} \ - \ \sum_{k = 1}^{n} \ \dfrac{1}{2k - 1} \bigg\rvert \ = \ \dfrac{1}{2n + 1} \ < \ \dfrac{1}{n},$$

but $(x_n)$ diverges.

Answer 4

Consider $x_n=\left|(\ln(n)\bmod 2)-1\right|$. The difference between consecutive square roots is small enough, the mod and abs "fold" the values into the interval $[0,1]$ (without increasing these distances), and in this interval the sequence is dense.