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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
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.
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.