Definition 8.3.1. Harmonic Series.
We call the series
\begin{equation*}
\sum_{k=1}^\infty \frac{1}{k}
\end{equation*}
the Harmonic Series.
Section 8.3 The Divergence Test and the Harmonic Series
Subsection The Relationship Between a Sequence and Series
We have looked at both infinite sequences and infinite series so far, and, to make things complicated, we’re really thinking about an infinite series (of terms from an infinite sequence) as an infinite sequence (of partial sums of the series). We’ve looked at how to visualize these (in both Graphing Sequences and Visualizing the Sequence of Partial Sums).
Let’s first start with defining a new series. This is a relatively important one by itself (it does have its own name), but it’s mostly an important series because it leads us into some new and interesting ways of thinking about series in general.
You might not recognize this, but we’ve worked with a version of this before. The example series that we plotted in Figure 8.2.5 was
\begin{equation*}
\sum_{k=0}^{\infty} \frac{3}{k+1}\text{.}
\end{equation*}
We can notice that if we re-index this by starting at \(k=1\) instead of \(k=0\text{,}\) we were really just looking at a scaled version of the harmonic series.
\begin{align*}
\sum_{k=0}^{\infty} \frac{3}{k+1} \amp = \sum_{k=1}^{\infty} \frac{3}{k}\\
\amp = 3\left(\sum_{k=1}^{\infty}\frac{1}{k}\right)
\end{align*}
Activity 8.3.1. Investigating the Harmonic Series.
(a)
Write out the first several terms of the harmonic series, terms from \(\left\{\dfrac{1}{k}\right\}_{k=1}^\infty\text{.}\) Write however many you need to get a feel for how the terms work.
(b)
Can you find out how many terms you would have to go “into” the series before the term was less than 0.00000001?
(c)
Can you do this same kind of thing, no matter how small? For instance, how many terms would you have to go into the series before the term was less than some real number \(\varepsilon\) where \(\varepsilon\gt 0\text{?}\)
(d)
Remind/explain/convince yourself that what we’ve really done is show that \(\displaystyle\lim_{k\to\infty}\frac{1}{k} =0\text{.}\) This isn’t a new or terribly interesting fact, but make sure that you understand why the argument above shows this.
(e)
Let’s do something very similar, but with \(\displaystyle\left\{\sum_{k=1}^n \frac{1}{k}\right\}_{n=1}^\infty\text{,}\) the sequence of partial sums, instead. Write out the first few partial sums. There’s no specific number that you need to write, but make sure to write enough partial sums to get a feel for how the partial sums work.
(f)
Can you find out how many terms you need to add up until the partial sum is larger than 1?
(g)
Can you find out how many terms you need to add up until the partial sum is larger than 5?
Hint.
(h)
Can you find out how many terms you need to add up until the partial sum is larger than 10?
Hint.
(i)
Do you think that for any positive number \(S\text{,}\) we can always find some partial sum \(\displaystyle\sum_{k=1}^n \frac{1}{k} \gt S\text{?}\) What do you think this would mean about
\begin{equation*}
\lim_{n\to\infty}S_n = \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k}\text{?}
\end{equation*}
To actually show that for any \(S \gt 0\) we could always find an \(n\gt 1\) where
\begin{equation*}
\sum_{k=1}^n \frac{1}{k} \gt S
\end{equation*}
is an extremely difficult task! We will show that the Harmonic Series diverges in a different way, but for now I want us to notice these contradictory results: we have a series whose terms get small, but whose partial sums do not seem to converge.
We have \(\dfrac{1}{k}\to 0\) but it seems like \(\displaystyle\lim_{n\to\infty} S_n\) does not exist. Is this behavior special to the Harmonic Series? Is this something we should make note of? Is there some other connection between the terms of a series and the behavior of the partial sums of the series that we need to note?
Let’s continue to think about this strange series, but actually prove that the series itself diverges.
Theorem 8.3.2. The Harmonic Series Diverges.
The Harmonic Series,
\begin{equation*}
\sum_{k=1}^\infty \frac{1}{k}\text{,}
\end{equation*}
diverges.
Proof.
Let’s assume, for the sake of eventual contradiction, that the harmonic series converges. Our goal in this proof is to show that this assumption (convergence) logically leads to an internal contradiction. This would mean that the assumption (convergence) cannot be true.
So, let’s assume that the harmonic series converges.
Based on our definition of series convergence (Definition 8.2.2), there exists some real number \(S\) such that:
\begin{equation*}
\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k} = S\text{.}
\end{equation*}
We’re going to think about this number, \(S\text{,}\) and show that there cannot be such a number.
First, let’s write out what \(S\) is:
\begin{equation*}
S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ...
\end{equation*}
We’re now going to systematically change the numbers being added together in order to create some number that is smaller than \(S\text{:}\) we’re going to take all of the odd terms and make them as small as the next term after it:
\begin{align*}
S \amp = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + ... \\
S \amp \gt \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{6}+ \frac{1}{6} + ...
\end{align*}
Note, though, that we can group together these duplicate terms and add them. Let’s do that!
\begin{align*}
S \amp \gt \left(\frac{1}{2} + \frac{1}{2}\right) + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{6}+ \frac{1}{6}\right) + ...\\
S \amp \gt 1 + \frac{1}{2} + \frac{1}{3} + ...
\end{align*}
But we should recognize this new series that is smaller than \(S\text{...}\)it’s the harmonic series! Which, by our initial assumption, is also \(S\text{!}\)
Ok, so what we have shown is that if the harmonic series converges, then it converges to some number \(S\) that has the contradictory property of being smaller than itself.
There is no such number.
This is a contradiction, then, and so the harmonic series must diverge.
This is a strange result, and one that has been brought up again and again by mathematicians throughout history. We’ll see that this series is notable because of its use later on in this chapter, but for now we can simply note that it is strange to see a series of terms that get so small (and so quickly) and yet the sum of those terms diverges.
The connection between the terms of a series and the behavior of the infinite series itself is maybe more mysterious than we initially thought. Since we will likely not have “access” to the formula for the partial sums (Note 8.2.6), we will want to explore these kinds of connections as much as we can. They will be the things to help us analyze an infinite series.
Subsection The Divergence Test
Theorem 8.3.3. Divergence Test.
For an infinite series \(\displaystyle\sum_{k=0}^\infty a_k\text{,}\) if the infinite series converges then
\begin{equation*}
\lim_{k\to\infty} a_k = 0\text{.}
\end{equation*}
This is equivalent to saying that if
\begin{equation*}
\lim_{k\to\infty} a_k\neq 0
\end{equation*}
then the infinite series \(\displaystyle\sum_{k=0}^\infty a_k\) diverges.
Proof.
We will prove the claim that if an infinite series converges, then its sequence of terms must converge to \(0\text{.}\)
This result will fall out of a simple exploration of what partial sums are. We noted in Section 8.2 that we can write any partial sum as the sum of the previous partial sum and the next term:
\begin{align*}
S_0 \amp = a_0 \\
S_1 \amp = a_0 + a_1 \\
\amp = S_0+ a_1\\
S_2 \amp = a_0+a_1+a_2\\
\amp = S_1 + a_2\\
S_3 \amp = a_0+a_1+a_2+a_3\\
\amp = S_2 + a_3\\
\amp \vdots \\
S_n \amp =a_0+a_1+...+a_n \\
\amp = S_{n-1}+a_n
\end{align*}
Let’s now say that the series we are dealing with converges. This means that \(\displaystyle\lim_{n\to\infty} S_n = S\) for some real number \(S\text{.}\)
It has to also be \(S\text{!}\) If the partial sums converge, then these two partial sums must converge to each other \(n\) increases:
\begin{equation*}
\lim_{n\to\infty} S_{n-1} = \lim_{n\to\infty} S_n \text{.}
\end{equation*}
So, since \(S_n = S_{n-1} + a_n\text{,}\) we can investigate the limit of \(a_n\text{:}\)
\begin{align*}
a_n \amp = S_n - S_{n-1} \\
\lim_{n\to\infty}a_n \amp = \lim_{n\to\infty}(S_n - S_{n-1}) \\
\amp = 0
\end{align*}
So of course the \(n\)th term has to converge to 0 in the limit!
Example 8.3.4.
Apply the Divergence Test to the following series and interpret the results.
(a)
\(\displaystyle \sum_{k=0}^\infty \frac{k^{15} - 4k^{10} + 10k^4}{e^{2k}}\)
Hint.
We can do a couple of things here! There is a nice result about limits of polynomials that we can use in the numerator (Polynomial End Behavior Limits). We could also get this same result using some other techniques, like what we use to prove that theorem. Or we can use L’Hôpital’s Rule to evaluate the limit, since we have a \(\frac{\infty}{\infty}\) indeterminate form.
(b)
\(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2+1}\)
Hint.
These terms are strange! The \((-1)^{k+1}\) part really just impacts the sign of the terms, since it is either \(1\) or \(-1\) depending on if \(k\) is even or odd.
We can consider only one sign (maybe the positive), and then try to make a conclusion about the alternating terms. Do they go to 0?
(c)
\(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sqrt[k]{k}}\)
Hint.
This is similar: focus on only the positive terms for now. But that denominator is also strange! If you want to focus only on the denominator, you can use the following friendly rearrangement:
\begin{align*}
\lim_{k\to\infty} \sqrt[k]{k}\amp =\lim_{k\to\infty} k^{\frac{1}{k}}\\
\amp =\lim_{k\to\infty} e^{\ln\left(k^{1/k}\right)}\\
\amp = \lim_{k\to\infty} e^{\frac{1}{k}\ln(k)}\\
\amp = e^{\lim_{k\to\infty} \frac{\ln(k)}{k}}
\end{align*}
Now you can use L’Hôpital’s Rule to evaluate this part!
Subsection Squeeze Theorem and Growth Rates
In order for us to calculate the limits of the terms of an infinite series, we are often using results like The Squeeze Theorem and L’Hôpital’s Rule. Let’s observe two facts that will help us out.
We don’t need to apply the Squeeze Theorem in “full,” since we’re only concerned with whether the limit is 0 or not. Let’s clarify with an example:
\begin{equation*}
\sum_{k=1}^\infty \frac{(-1)^{k+1}2k^2}{7k^2+k+1} = \frac{2}{9} - \frac{8}{31} + \frac{18}{67} - \frac{32}{117} +-...
\end{equation*}
For us to apply the Divergence Test, we really should note that:
\begin{equation*}
-\frac{2k^2}{7k^2+k+1}\leq \frac{(-1)^{k+1}2k^2}{7k^2+k+1} \leq \frac{2k^2}{7k^2+k+1}\text{.}
\end{equation*}
We could look at the limits of both of the bounds, but since we’re concerned with whether they are 0 or not, we can simply check one:
\begin{equation*}
\lim_{k\to\infty} \left(\frac{2k^2}{7k^2+k+1}\right) = \frac{2}{7}\text{.}
\end{equation*}
Since we know that the negative version of this limit isn’t 0 either, we can see that the limit \(\displaystyle\lim_{k\to\infty}\frac{(-1)^{k+1}2k^2}{7k^2+k+1}\) doesn’t exist. Thus, the series diverges.
We can test just the positive version of the limit in each of these cases, then!
" Lastly, we can notice now that L’Hôpital’s Rule can be difficult to apply, especially since the functions we’re working with are not required to be differentiable. For instance, how could we consider the series
\begin{equation*}
\sum_{k=0}^\infty \frac{k!}{3^k}
\end{equation*}
by thinking about the limit
\begin{equation*}
\lim_{k\to\infty} \frac{k!}{3^k}\text{?}
\end{equation*}
Because we can’t think about the derivative of \(k!\) (since the function is not continuous, and thus not differentiable in the traditional sense), we need to consider this limit in a different way. We can instead appeal to growth rates of these functions: that is, the rate at which these functions approach infinity.
Growth Rates of Functions.
If \(m\) and \(n\) are real numbers with \(0\lt m\lt n\text{,}\) then we can arrange the growth rates of the following functions from “slow” to “fast:”
\begin{equation*}
\ln(k)\lt\lt k^m \lt\lt k^m\ln(k) \lt\lt k^n \lt \lt m^k \lt \lt n^k \lt \lt k! \lt\lt k^k
\end{equation*}
So, returning to our series \(\displaystyle \sum_{k=0}^\infty \frac{k!}{3^k}\text{,}\) we can see that
\begin{equation*}
\lim_{k\to\infty} \frac{k!}{3^k} = \infty
\end{equation*}
since \(k!\) has a “faster” growth rate than \(3^k\text{.}\)
Note 8.3.5.
It’s worth investigating this yourself! Try to convince yourself that you understand why exponentials grow “slower” than factorials. Why does \(k^k\) have a “faster” growth rate than \(k!\text{?}\)
Practice Problems Practice Problems
1.
Give an example of a series that diverges due to the Divergence Test. Explain.
2.
Give an example of a series that fails the Divergence Test. Do you know if this series converges or diverges? Explain.
3.
Apply the Divergence Test to each of the following series. Explain the conclusions of the test for each series.
(a)
\(\displaystyle\sum_{k=1}^\infty \left(\frac{k}{k^2+1}\right)\)
(b)
\(\displaystyle\sum_{k=1}^\infty \left(\frac{k}{e^k}\right)\)
(c)
\(\displaystyle\sum_{k=2}^\infty \left( \frac{1}{k\ln(k)} \right)\)
(d)
\(\displaystyle\sum_{k=2}^\infty \left(\frac{k}{\ln(k)}\right)\)
(e)
\(\displaystyle\sum_{k=1}^\infty \left(\sqrt[k]{k}\right)\)
(f)
\(\displaystyle\sum_{k=0}^\infty \left( \frac{k^2+k}{e^k} \right)\)
(g)
\(\displaystyle\sum_{k=1}^\infty \left( \frac{1}{k^2+1} \right)\)
(h)
\(\displaystyle\sum_{k=1}^\infty \left( \frac{k!}{k^2} \right)\)
(i)
\(\displaystyle\sum_{k=1}^\infty \left( \frac{(-1)^k(k+1)}{k+\frac{1}{k}} \right)\)
