The comparison test is an easy test for convergence or divergence when we know that, term by term, the terms of a test series are either greater than or less than those of our series of interest.
Here is a statement of the test in two parts, then we'll do some examples to illustrate it:
Here is the logic behind the first statement of the comparison test:
We get the convergent series used for comparison from our growing set of convergent series. For example, it might be a convergent geometric series or a convergent p-series.
The logic behind the second part of the comparison test goes like this:
We get the known divergent series used for comparison from among the many divergent series we've seen. For example, it might be a divergent geometric series (r < 1) or a divergent p-series (p < 1).
Solution: When we look at this series, it's how rapidly the denominator grows that's important. We know that the series
the harmonic series, diverges, but that the p-series
converges. Now if each term of the series of interest is greater than 1/n for n > 1 then the series diverges, and if each term is less than 1/n2, then the series converges by the comparison test.
We know that for n ≥ 4, n! > n2, so each term of the series with an = 1/n! is less than the corresponding term of our convergent p-series, therefore the series converges. Here's a graphical look at these three series, 1/n in green, 1/n2 in magenta and 1/n! in blue. Notice that for n ≥ 4, 1/n! is smaller, term-by-term, than 1/n2.
Now it's worth going back to the first green box above. The comparison test states that the terms of our test series (1/n!) must only be less, term-by-term, than a known convergent series for n > N, where N is some integer. In this example, N = 3. For all n > 3, 1/n! < 1/n2, so the series converges. The extra stuff at the "front end" of the series (n = 1, 2, 3) is just a constant added on to a convergent series.
Solution: For a series like this, it's helpful to reduce it by approximation. In the first approximation, the 5 in the denominator doesn't really matter that much for large n, so let's drop it to make an approximation of the series:
Now the series with terms an = 1/n3/2 is a convergent p-series (p = 3/2 > 1). That suggests that we can show that each of the terms of our series is smaller than the corresponding term of the p-series with p = 3/2. First we set up the trial inequality:
Cross multiplication gives:
Using the laws of exponents on the left, we get:
which reduces to
Because 0 < 5 is always true, our series is, term-by-term, smaller than the convergent p-series with an = 1/n3/2, so it converges.
Here (right column, top) is a table of the first ten terms of our series and the comparison series:
Below is a graphical representation of that comparison out to n = 30. While our the size of the terms of our function get closer to those of the comparison function, our algebra proves that they never cross.
Solution: This series contains an oscillating factor, cos(1/n). Sometimes that factor is positive and sometimes it's negative. Now 1/n approaches zero as n gets large, so the cosine part approaches 1. The important thing is that it won't ever exceed 1.
That means that this function is never greater than the convergent p-series with terms an = 1/n2, so the series converges by the comparison test.
Easy peasy, lemon-squeezy.
Solution: It's not too difficult to see that this series has to diverge. If we drop the 1 in the numerator, each term is just n1/2, and the sum will grow infinitely large. But let's do it by comparison with the divergent p-series with terms an = 1/n1/2.
First let's break the sum into two parts.
Now the second is the divergent p-series with terms an = 1/n1/2 so the n1/2 term just adds an ever-increasing amount to an already divergent series, therefore the series diverges.
There are a couple of things you should be aware of when using the comparison test.
Determine whether these series converg using the comparison test.
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