Prove Guass’ test for the convergence of a series with for all . This test says that if there is an , an , and an such that
where for all , then the series converges if and diverges if .
Proof. Let be a series of positive terms, and assume there exists an , an , and an such that
for all , and for all . Now we consider the three cases , and .
Case 1. Assume . Then we have
Since for all we have
But then since we have and so for all sufficiently large we have
(since we can make arbitrarily small so even if it is smaller in absolute value than and so must be positive). Since this term is positive we then have
for all sufficiently large . By Raabe’s test (Section 10.16, Exercise #16) we then have diverges.
Case 2. Assume . Then we have
for (since and ). Hence, by Raabe’s test again we have converges.
Case 3. Assume . From a previous exercise (Section 10.16, Exercise #15) we know that if is a sequence of positive terms, is a sequence of positive terms such that diverges and where
then diverges. So, in this case, let . Then we know from Example 2 on page 398 of Apostol that diverges. So, we let
But then, since the logarithm is an increasing function, we know . Furthermore,
since for all sufficiently large , and by equation (10.11) on page 380 of Apostol. Therefore, for sufficiently large we have
Hence, by the previous exercise, we have diverges.
So, putting all three cases together we have the requested result, converges if and diverges if