The last convergence tool we have is the comparison test. If all else fails, we should compare our Pandora's box to another one. If we look at the other one, and we decide the other one is bursting at the seems, we know it's safe to open ours up.
We studied improper integrals a while back, and we learned that, if f ≤ g on the interval (c,∞), then

and

Here, when f and g are nonnegative, there's a smaller area under f and than under g. If the big area is finite, the smaller area must be finite too. If the small area is infinite, the bigger area must be infinite. If the other box is smaller and it is too dangerous to open, then ours is too. If is the other box is larger and it is safe to open, so is ours.
Since we can visualize a series as an area , we can use the same intuition to compare series. Suppose we have two series

where
0 ≤ an ≤ bn
for all n. Then the small area described by the series a is contained in the big area described by the series B.
This tells us two useful things:
- If the area described by B is finite, the smaller area described by a must also be finite.
- If the area described by a is infinite, the bigger area described by B must also be infinite.
Be Careful: As with improper integrals, we have to be careful about which way the comparisons go.
- If the area described by a is finite, that doesn't tell us anything useful. The area described by B could be a larger finite value, or it could be infinite.
- If the area described by B is infinite, that doesn't tell us anything useful. The smaller area described by a could be either finite or infinite.
The tricky part, as with improper integrals, is finding the correct series to compare with. We can't compare gremlins to grilled cheese. We need something similar and easy to tell if the series converge or diverge. We like to use series of the form

whenever possible, since we can easily tell whether such series diverge or converge.
Be Careful: We can't use the comparison test if we can't find something to compare with. For example, we can't use the comparison test on

The only thing we can see to compare the term with is
, but the inequality goes the wrong way:

Practice:
Use the comparison test to determine whether the series  converges or diverges. | |
The given series looks like the series 
which converges, so we'll guess that the given series converges too. In order to convince the teacher, we have to find a series Σ bn that converges and has bigger terms than the given series. Since making the denominator smaller makes the whole fraction bigger, it follows that 
Then since the series 
converges and has bigger terms than the original series, the comparison test says that the original series  must converge also. Since limits of summation don't affect whether a series converges, it's okay if the relationship 0 < an ≤ bn doesn't hold for all the terms at the beginning, so long as it holds for all the terms from some point onwards. | |
Use the comparison test to determine if the series 
converges or diverges. | |
The terms look roughly like or , so we'll guess that this series diverges. To be convincing, we need to find a series with smaller terms whose sum diverges. Since n2 – 81 < n2 it follows that 
Since diverges and has smaller terms than the series we were given, the series we were given diverges. In the previous example, we didn't have the relationship 
for all n. The terms are negative for 1 ≤ n < 9, and not even defined for n = 9. However, the relationship does hold for n ≥ 10, and that's good enough. There's one down side to the comparison test. When using it, sometimes we have to use other tests also to show the convergence or divergence of the series we're comparing to. Because of this, the comparison test is meant to be a last resort. | |
Does the series converge or diverge? | |
If we make the denominator smaller we make the fraction bigger, so we could try either of these comparisons: 
Comparing to isn't helpful. The harmonic series diverges, but that doesn't tell us anything about series with smaller terms. Comparing to is helpful, though. The series converges because it's geometric with ratio 
Since the given series has smaller terms, it has to converge too. Here's one of our favorite tricks to use with the comparison test. It makes perfect sense once you see it, but it's sort of hard to think of the first time. | |
Does the series 
converge or diverge? | |
Since sin n is always between -1 and 1, (2 + sin n) is always between 1 and 3. That means we can say 
We know that 
converges by the p-test . Multiplying the terms by a constant doesn't affect convergence , so 
converges too. Since 
the comparison test says that the original series must converge. | |
Use the comparison test to determine whether the series converges or diverges.

Answer
This series looks like
, so we guess it diverges. Since making the denominator smaller makes the fraction bigger,

(for n > 5, anyway). Since the harmonic series diverges and the series

has bigger terms, this series has to diverge too. The comparison test tells us so.
Use the comparison test to determine whether the series converges or diverges.
Answer
For n ≥ 2 we know that

The series

converges because it's geometric with
. Since the series

has smaller terms, the comparison test says this series must converge also.
Use the comparison test to determine whether the series converges or diverges.

Answer
Making the denominator smaller makes the fraction bigger, so

The series

is geometric with ratio
, so it converges. The comparison test says the series

which has smaller terms, has to converge too.
Use the comparison test to determine whether the series converges or diverges.

Hint
for n ≥ 1
Answer
This one's a little sneaky. Do the terms look like
or like
? Either way, we guess the series diverges. To show this we need to find a divergent series with smaller terms. If we make the denominator bigger we make the fraction smaller, so replace
with n:

Since multiplication by a constant doesn't affect whether the series diverges,

diverges. The comparison test says that

which has bigger terms, also diverges.
Use the comparison test to determine whether the series converges or diverges.

Answer
This kind-of-sort-of looks like
, which converges, guess the series converges. We need to find a convergent series with bigger terms. To make a fraction bigger we can make the denominator smaller and/or make the numerator bigger.
To make the denominator smaller, use the fact that n < en to replace en with n:

Then make the numerator bigger too. Since cos n is always between -1 and 1,
2 – cos n ≤ 3.
This means

We know that

converges because of the p-test and because multiplication by a constant doesn't affect whether the series diverges. This means the series with smaller terms,

has to converge.