Use the alternating series test.
converges if and all the are non-negative.
For your example, .
Can you finish..? (note that your example starts with n=1 not n=0 as in the definition but it won't make a difference).
Indeed, if denotes the sequence of partial sums of the initial series, the second series has partial sums . If (i.e. if the first series converges) then as well (as a subsequence). And if , then and as well, hence classically .
Thus, it suffices to prove the convergence of the series where the terms are grouped two by two. And, as a matter of fact, this new series converges absolutely, as we will prove.
A short computation shows as , hence . Substituting shows that the absolute value of the general term of the new series is asymptotically equivalent to . Since , this series converges (usual Riemann series). Thus the new series converges absolutely, hence it converges, and the inital series as well.