The comparison test

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We compare infinite series to each other using inequalities.

If $0 \leq a_k \leq b_k$ for all $k$ , which of the following do you think are true?

If $\sum ^\infty b_k$ converges then $\sum ^\infty a_k$ converges If $\sum ^\infty a_k$ converges then $\sum ^\infty b_k$ converges If $\sum ^\infty b_k$ diverges then $\sum ^\infty a_k$ diverges If $\sum ^\infty a_k$ diverges then $\sum ^\infty b_k$ diverges

If you got this question right, you already understand the comparison test!

The Comparison Test Let $\sum _{k=0}^\infty a_k$ and $\sum _{k=0}^\infty b_k$ be series with positive terms:

If $a_k \leq b_k$ and $\sum a_k$ is divergent, then $\sum b_k$ is divergent.
If $a_k \leq b_k$ and $\sum b_k$ is convergent, then $\sum a_k$ 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.

Is $\sum _{k=1}^\infty \frac {k}{7+k^3}$ convergent or divergent? Justify your answer using the comparison test.

Intuitively, we should feel that this should be similar to the series $\sum _{k=1}^\infty \frac {k}{7+ k^3} \approx \sum _{k=1}^\infty \frac {k}{0+k^3}$ since the additional term of $\answer [given]{7}$ in the denominator should not matter when $k$ is large. Since $\sum _{k=1}^\infty \frac {k}{k^3} =\sum _{k=1}^\infty \frac {1}{k^2}$ we see that our original sum $\sum _{k=1}^\infty \frac {k}{7+ k^3} \approx \sum _{k=1}^\infty \frac {1}{k^2}.$ We know $\sum _{k=1}^\infty \frac {1}{k^2}$ is convergent by the $p$ -series test (which ultimately comes from the integral test), so we guess that our series in question is also convergent. Since we want to prove convergence, we need find a series whose terms are always larger than $\frac {k}{7+ k^3}$ . Note $\frac {k}{1+k^3} < \frac {k}{0+k^3}$ since we have made the denominator largersmaller , which makes the fraction largersmaller . Thus $\frac {k}{1+k^3} < \frac {1}{k^2}$ . Since $\sum _{k=1}^\infty \frac {1}{k^2}$ converges, then our original series also converges by the comparison test.

Is $\sum _{k=1}^\infty \frac {k^2}{5+k^3}$ convergent or divergent? Justify your answer using the comparison Test.

Intuitively, we should feel that this should be similar to the series $\sum _{k=1}^\infty \frac {k^2}{5+k^3} \approx \sum _{k=1}^\infty \frac {k^2}{0+k^3}$ since the additional term of $\answer [given]{5}$ in the denominator should not matter when $k$ is large. Since $\sum _{k=1}^\infty \frac {k^2}{k^3} =\sum _{k=1}^\infty \frac {1}{k}$ we see that our original sum $\sum _{k=1}^\infty \frac {k^2}{5+ k^3} \approx \sum _{k=1}^\infty \frac {1}{k}.$ We know $\sum ^\infty \frac {1}{k}$ is divergent by the $p$ -series test (which ultimately comes from the integral test), so we guess that our series in question is divergent. Since we want to prove divergence, we need to find a series whose terms are always smaller than $\frac {k^2}{5+ k^3}$ . We can’t use $\frac {1}{k}$ , because those terms are larger. Hence, this requires some creativity. Write with me. $\frac {k^2}{5+k^3} \geq \frac {k^2}{5k^3+k^3}$ since we have made the denominator largersmaller , which makes the fraction largersmaller , note that $k^3 \geq 1$ whenever $k \geq 1$ . Thus $\frac {k}{1+k^3} \geq \frac {1}{6k}$ . Since $\sum _{k=1}^\infty \frac {1}{6k} = \frac {1}{6} \sum _{k=1}^\infty \frac {1}{k}$ diverges, then our original series also diverges by the comparison test.