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).
Fine. Your series is . I say: this series converges if and only if the series (the terms are inside parentheses) converges.
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.