Limit comparison test
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Suppose that we have two series and with for all .
Then if with then either both series converge or both series diverge.
Because we know that for all there is a positive integer such that for all we have that , or equivalently
As we can choose to be sufficiently small such that is positive. So and by the direct comparison test, if converges then so does .
Similarly , so if converges, again by the direct comparison test, so does .
That is both series converge or both series diverge.
We want to determine if the series converges. For this we compare with the convergent series .
As we have that the original series also converges.
One can state a one-sided comparison test by using limit superior. Let for all . Then if with and converges, necessarily converges.
Let and for all natural numbers . Now does not exist, so we cannot apply the standard comparison test. However, and since converges, the one-sided comparison test implies that converges.
Converse of the one-sided comparison test
Let for all . If diverges and converges, then necessarily , that is, . The essential content here is that in some sense the numbers are larger than the numbers .
Let be analytic in the unit disc and have image of finite area. By Parseval's formula the area of the image of is . Moreover, diverges. Therefore by the converse of the comparison test, we have , that is, .