Question about bounded subsequences Let $(x_n)$ be a sequence of real numbers satisfying the following:
Every bounded subsequence $(x_{n_j})$ of $(x_n)$ is convergent.
Prove:

*

*Is possible to $(x_n)$ have convergent subsequences but $(x_n)$ to be not convergent.

*If $(x_n)$ is bounded, then it is convergent.

Number (2) follows from the property because every sequence is a subsequence of itself. For number (1) I do not know if it is enough to find an example or if I have to prove it in general.
thank you

As "unknown" has pointed out, $\big((-1)^n\big)_{n=1}^\infty$ has convergent subsequences and does not converge.
The following is a sequence that does not converge but in which every bounded subsequence (and it has many bounded subsequences) converges to the same limit:
$$
1,2,1,3,1,4,1,5,1,6,1,7,1,8,\ldots
$$
So this has convergence subsequences although it does not converge. For example:
$$
\underbrace{1,2,1,3,1,4,1,5,1,6,\ldots,1,500\,000}_\text{the first one million terms},\,\,\underbrace{1,1,1,1,\ldots\ldots\ldots}_{\begin{smallmatrix} \text{all later terms} \\ \text{whose indices are} \\ \text{multiples of 7}\end{smallmatrix}}
$$
This is a convergent subsequence.
