Recall that a sequence [tex]x_n[/tex] is convergent if and only if [tex]x_n[/tex] is also a Cauchy sequence, which means to say that for any [tex]\varepsilon>0[/tex], we can find a sufficiently large [tex]N[/tex] for which
[tex]|x_m-x_n|<\varepsilon[/tex]
whenever both [tex]m[/tex] and [tex]n[/tex] exceed [tex]N[/tex].
But this never happens if we choose [tex]m=n+1[/tex] and [tex]0<\varepsilon<2[/tex]; under these conditions, we have
[tex]|x_m-x_n|=|x_{n+1}-x_n|=2\not<2[/tex]
Therefore [tex]x_n[/tex] is not a Cauchy sequence and hence does not converge.
