notice that by the Alternating Series Test, (−1)n /nγ
converges for all γ > 0 whereas 1/n2γ converges if and only if γ > 1/2. Finally,
by setting an = (−1)n /nγ − 3/4n2γ and cn = (−1)n /nγ − 1/4n2γ, an application of
our Theorem enables us to conclude that