A [n] = (- 1) n, ∑ {1≤ n} (a [2n-1]+a [2n]) converges, but ∑{ 1 ≤ n} a[n] diverges.

If the condition a[n] ≥ 0 is added, their convergence is equivalent.

This limit does exist.

But I guess F (x) = x (4/3) sin (1/x) takes derivatives everywhere but the derivative is discontinuous at x = 0.

In this case, your derivative is wrong.

For x ≠ 0, f' (x) = 4/3 x (1/3) sin (1/x)-x (-2/3) cos (1/x).

This is unbounded in any neighborhood of 0.