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Suppose and . Then . This trivial observation turns out to be the root of an incredibly powerful method for testing the convergence or divergence of infinite series.
An obvious generalization of the above is that if we have two infinite sequences satisfying for all , then for all . Suppose in addition that for all , and that converges to its sum . Then . However, the partial sums for are non-negative and bounded from above. Therefore, by an earlier theorem (this probably needs to be filled out in an earlier topic!), converges as well. Intuitively, if is "small enough" such that converges, and is smaller than , then is also "small enough" so that converges as well.
Similarly, if , but this time diverges, then also diverges as well. Indeed, must diverge, because if it converged, then so would by what we have just shown, a contradiction. Therefore, we have the following theorem:
Theorem (Direct Comparison Test). Suppose for all .
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If converges, then so does .
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If diverges, then so does .
Exercise 1. Show that the requirement that be non-negative in the direct comparison test is essential. That is, exhibit a convergent series such that but diverges.
Exercise 2. Determine the convergence or divergence of the series
Notice that if we want to use this test to show that a particular series converges or diverges, we need to be able to compare that series to some reference series whose convergence or divergence we already know. And what general series do we know converge or diverge? Geometric series! Throughout the remainder of the sections on infinite series, we will keep returning to comparisons against geometric series, or theorems which are proven via comparison against geometric series, over and over.
(There's more stuff to write here: actual concrete examples/exercises, and the limit comparison test. I'll do it later.)