We compare infinite series to each other using inequalities.

**already**understand the

*comparison test*!

- If and is divergent, then is divergent.
- If and is convergent, then is convergent.

Notice that this test, just like the root and ratio tests, require us to have positive terms in our series. Of course, what we really mean by positive terms is that a series should eventually have only positive terms. As always, a finite number of negative terms doesn’t affect the overall convergence or divergence of a series.

The proof of this theorem is beyond this course, but it should make intuitive sense.

- Making the terms of a convergent series smaller should result in another convergent series.
- Making the terms of divergent series larger should result in another divergent series.

While this theorem is intuitive, its use involves considerable creativity. You have to:

- (a)
- Find a simpler series which “behaves like” your series.
- (b)
- Use this simpler series to predict whether the original series converges or diverges.
- (c)
- If you predict your series is convergent, you have to hunt for a simpler series which is also convergent, but all of whose terms are larger.
- (d)
- If you predict your series is divergent, you have to hunt for a simpler series which is also divergent, but all of whose terms are smaller.

All of these steps (aside from the second) require real insight and creativity. This is not a “mechanical” test.

**larger**than . Note since we have made the denominator largersmaller , which makes the fraction largersmaller . Thus . Since converges, then our original series also converges by the comparison test.

**smaller**than . We can’t use , because those terms are larger. Hence,

**this requires some creativity.**Write with me. since we have made the denominator largersmaller , which makes the fraction largersmaller , note that whenever . Thus . Since diverges, then our original series also diverges by the comparison test.