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Mathematical Expression Editor
We compare infinite series to each other using limits.
Using the comparison test can be hard, because finding the right sequence of
inequalities is difficult. The limit comparison test eliminates this part of the
method.
The Limit Comparison Test Let \(\sum _{k=0}^\infty a_k\) and \(\sum _{k=0}^\infty b_k\) be series with positive terms and let
\[ \lim _{k \to \infty } \frac {a_k}{b_k} = L. \]
If \(0<L<\infty \) then either both series converge, or they both diverge.
If \(L=0\) and \(\sum b_k\) converges, then \(\sum a_k\) converges.
If \(L=\infty \) and \(\sum b_k\) diverges, then \(\sum a_k\) diverges.
This theorem should make intuitive sense.
If \(0<L<\infty \) then we have \(a_k \approx L \cdot b_k\) for large \(k\), so the behavior of the respective series should
be the same.
If \(L=0\) then \(a_k\) should be way less than \(b_k\). So if \(b_k\) converges, \(a_k\) should also converge
by the comparison test.
If \(L=\infty \), then \(a_k\) should be way greater than \(b_k\). So if \(b_k\) diverges, \(a_k\) should also diverge
by the comparison test.
The way we actually use this in practice still involves some creativity: we have to
decide on a “similar” series for which we know the convergence properties. However,
unlike the comparison test, we can just mechanically take a limit of the ratio of our
guess with our original series, instead of having to “get our hands dirty” with
inequalities.
Is \(\sum _{k=1}^\infty \frac {\ln (k)}{k^3}\) convergent or divergent? Justify your answer using the limit comparison test.
We
should expect that this series will converge, because \(\ln (k)\) goes to infinity slower than \(k\), so
the series is “no worse” than the \(p\)-series with \(p=2\). In the notation of the theorem, let
\[ a_k = \frac {\ln (k)}{k^3}. \]
We
will use the limit comparison test with the series
\[ \sum _{k=1}^\infty \frac {1}{k^2}, \]
so that
\[ b_k = \frac {1}{k^2}. \]
To apply the limit
comparison test, examine the limit \begin{align*} \lim _{k \to \infty } \frac {a_k}{b_k} &= \lim _{k \to \infty } \frac {\ln (k)}{k} \\ &= 0 \end{align*}
Since \(\sum _{k=1}^\infty b_k\) is convergent by the \(p\)-series test with \(p=2\), then the limit comparison test applies,
and \(\sum _{k=1}^\infty \frac {\ln (k)}{k^3}\) must also converge.
If the limit comparison test is easier to use than the comparison test, why do we even
have the comparison test? Sometimes, the comparison test is actually more powerful.
The next example illustrates this idea.
Consider \(\sum _{k=1}^\infty \frac {1+\sin (k)}{2^k}\). What happens when you try to use each of the comparison tests with \(\sum _{k=1}^\infty \frac {1}{2^k}\)?
The limit comparison test shows that the original series is convergent. The
limit comparison test shows that the original series is divergent. The limit
comparison test does not apply because the limit in question does not exist. The
comparison test can be used to show that the original series converges. The comparison test can be used to show that the original series diverges.
\(\frac {a_k}{b_k} = 1+\sin (k)\), which does not have a limit as \(k \to \infty \), so the limit comparison test does not apply.
On the other hand, we can see that
\[ 0<1+\sin (k)<2, \]
so
\[ \frac {1+\sin (k)}{2^k} < \frac {2}{2^k}, \]
which is a convergent geometric
series with \(r = \frac {1}{2}<1\). Thus the original series converges via the comparison test.
Let’s pause another moment to consider the task of choosing a test to use when
analyzing a series for convergence or divergence. Take a few minutes to make a list of
all the tests we know so far, and the best situations in which to use each of
them.