Suppose that $\{x_{n}\}$ is a sequence of real numbers satisfying the following. For every $\epsilon > 0$, there exists positive integer $n_{0}$ such that $\left |x_{n+1}-x_{n}\right |$<$\epsilon, \ \forall$ $n \geq n_{0}$. Then sequence ${x_{n}}$ is
1). Bounded but not necessarily Cauchy
2). Cauchy but not necessarily bounded
3). Convergent
4). Not necessarily bounded
Solution I tried:
According to the given condition in question this sequence is Cauchy,then option 1 and 2 can be rejected because in 1 it is not Cauchy and in 2 it is not bounded. I am confused that the answer is actually 4 how?
Please help
The sequence is not Cauchy. The Cauchy condition is far stronger than merely requiring differences of consecutive terms to be less than $\epsilon$!
To recap, the Cauchy condition requires $|x_m - x_n| < \epsilon$ for all $m, n > n_0$ and not just $|x_{n+1} - x_n| < \epsilon$ for all $n > n_0$. Notice: in the actual Cauchy condition, there is an extra integer $m$ variable whereas in your condition, $m$ is kept fixed at $n+1$. Unfortunately, keeping $m$ fixed at $n+1$ is too weak a condition to guarantee anything like convergence or even boundedness.
Indeed, consider $x_n = \sum\limits_{k = 1}^n \frac{1}{k}$ for $n \in \mathbb{Z}_+$. Then $|x_{n+1} - x_{n}| = \frac{1}{n+1}$. So for any given $\epsilon > 0$, you can find an integer $n_0 > \frac{1}{\epsilon}$ to guarantee that consecutive terms have differences less than $\epsilon$ for all integers $n > n_0$. But the $x_n$'s are partial sums of the harmonic series and hence they are not even bounded let alone Cauchy.