Series test for convergence/divergence

[parisse]

sum(1/n^alpha,n) is convergent if alpha>1, divergent for alpha<=1 (this is easy to prove by comparing with int(1/x^alpha,x)). If f(n) is equivalent to g(n) and g(n) has constant sign then sum(f(n),n) and sum(g(n),n) are both convergent or both divergent.
It's more complicated for non constant sign term, like sum((-1)^n/n^alpha,n) is convergent for alpha>0, but sum((-1)^n/f(n),n) may be divergent even if f(n) is equivalent to n^alpha, for this kind of series you will usually need a series expansion of f(n) at n=inf until the remainder of the expansion is O(1/n^alpha) with alpha>1. Like for example sum((-1)^n/sqrt(n+(-1)^n*sqrt(n)),n)