M344: Calculus III
Section 11.6
Absolute Convergence¶
Up to this point in our study, we've seen convergence tests for series with positive terms and for alternating series. But sometimes, we will have series where the signs of the terms oscillate irregularly. In these cases it is frequently useful to consider a corresponding series whose terms are all positive.
Given any series $\sum a_n$, its corresponding absolute series is
$$
\sum_{n=1}^{\infty} \lvert a_n \rvert = \lvert a_1 \rvert + \lvert a_2\rvert + \lvert a_3\rvert + \cdots.
$$
The terms of the absolute series are the absolute values of the terms of the original series.
Definition¶
A series $\sum a_n$ is called absolutely convergent if its corresponding absolute series, $\sum \lvert a_n \rvert$ is convergent.
Notice that in case $\sum a_n$ only has positive terms, i.e., if $\lvert a_n \rvert = a_n$, then absolute convergence is the same as convergence.
1. Example¶
The series
$$
\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n^2} = 1 - \dfrac{1}{2^2} + \dfrac{1}{3^2} - \dfrac{1}{4^2} + \cdots
$$
is absolutely convergent. To show this, we notice that $\lvert a_n \rvert = \dfrac{1}{n^2}$. Therefore the absolute series is
$$
\sum_{n=1}^{\infty} \left\lvert \dfrac{(-1)^{n-1}}{n^2}\right\rvert = \sum_{n=1}^{\infty} \dfrac{1}{n^2},
$$
which is a convergent $p$-series with $p$ = 2.
Definition¶
A series $\sum a_n$ is called conditionally convergent if it is convergent, but not absolutely convergent.
2.Example¶
Recall that the alternating harmonic series,
$$
\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots
$$
is convergent.
Implicit Exercise. If you do not "recall" this fact, either prove it again, or review the section of the notes where we proved it.
However, this same series is not absolutely convergent since its absolute series is
$$
\sum_{n=1}^{\infty} \left\lvert \dfrac{(-1)^{n-1}}{n}\right\rvert = \sum_{n=1}^{\infty} \dfrac{1}{n} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \cdots,
$$
which is the harmonic series. The harmonic series is an example of a $p$-series with $p = 1$, and thus is divergent. (Why?) Therefore the alternating harmonic series is conditionally convergent.
Theorem¶
If a series $\sum a_n$ is absolutely convergent, then it is convergent.
Proof
Let's consider $a_n + \lvert a_n \rvert$. Since $\lvert a_n \rvert$ is either equal to $a_n$ or $-a_n$, we have the following inequality.
$$ 0 \le a_n + \lvert a_n \rvert \le 2\lvert a_n \rvert $$
If $\sum a_n$ is absolutely convergent, then $\sum \lvert a_n \rvert$ is convergent, so $\sum 2\lvert a_n \rvert$ is also convergent. By the comparison test, $\sum \left( a_n + \lvert a_n \rvert \right)$ is convergent. Thus $\sum a_n$ is the difference of two convergent series, and is therefore also convergent.
$$ \sum a_n = \sum \left( a_n + \lvert a_n \rvert \right) - \sum \lvert a_n \rvert $$
1. Example¶
Determine where the series is absolutely convergent, conditionally convergent, or divergent.
$$
\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{\sqrt{n}}
$$
Check your answer
The series is conditionally convergent.
Video solution
2. Example¶
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n^{3}+1}
$$
Check your answer
The series is absolutely convergent.
Video solution
3. Example¶
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
$$
\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{n}{n^{2}+4}
$$
Check your answer
The series is conditionally convergent.
Video solution
The Ratio Test¶
One of the most powerful and versitile convergence tests is called the ratio test. In this test, we investigate the behavior of the ratio of adjacent terms in the series. If they are substantially decreasing, then the series is absolutely convergent.
Theorem¶
If $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = L < 1$, then the series $\displaystyle \sum_{n=1}^{\infty} a_n$ is absolutely convergent (and therefore convergent).
If $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = L > 1$ or $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = \infty$ then the series $\displaystyle \sum_{n=1}^{\infty} a_n$ is divergent.
If $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = 1$, then the Ratio Test is inconclusive.
Note. Case 3 of this theorem is inconclusive because it is possible that a series could be either convergent or divergent while satisfying this property. The fundamental illustrative examples of this fact are our friends the harmonic and alternating harmonic series. (Implicit Exercise: Test them!)
When we find ourselves mired in case 3, we need to do more (other) work, and try different convergence tests on our series.
Video Proof
Examples¶
We now consider some examples, starting with two that fall under case 3.
4. Example¶
Apply the ratio test to the series
$$
\sum_{n=1}^{\infty} \dfrac{1}{n^2}.
$$
Solution
Computing the limit $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert$, we get
$$ \lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = \lim\limits_{n\to\infty} \left\lvert \dfrac{\dfrac{1}{(n+1)^2}}{\dfrac{1}{n^2}} \right\rvert = \lim\limits_{n\to\infty} \dfrac{n^2}{(n+1)^2} = 1. $$
Therefore the Ratio Test does not apply to this series. However, we know that $\sum \dfrac{1}{n^2}$ is a convergent $p$-series. Since the terms are all positive to begin with, this series is absolutely convergent.
5. Example¶
Apply the Ratio Test to the series
$$
\sum_{n=1}^{\infty} = \dfrac{1}{\sqrt{n}}.
$$
Solution
Computing the limit $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert$, we get
$$ \lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = \lim\limits_{n\to\infty} \left\lvert \dfrac{1}{\sqrt{n+1}} \dfrac{\sqrt{n}}{1} \right\rvert = \lim\limits_{n\to\infty} \dfrac{\sqrt{n}}{\sqrt{n+1}} = 1. $$
Therefore the Ratio Test does not apply to this series. By the $p$-Test, this series is divergent.
The previous examples should help to cement the idea that if $\lim\limits_{n\to\infty} \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert = 1$, the Ratio Test is not applicable, and we must use a different test to determine if the series $\sum a_n$ converges or diverges.
6. Example¶
Use the Ratio Test to determine whether the series is convergent or divergent.
$$
\sum_{n=0}^{\infty}\dfrac{(-3)^{n}}{(2n+1)!}
$$
Check your answer
The series is absolutely convergent.
Video solution
7. Example¶
Use the Ratio Test to determine whether the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{n!}{n^{n}}
$$
Check your answer
The series is absolutely convergent.
Video solution
8. Example¶
Use the Ratio Test to determine whether the series is convergent or divergent.
$$
\dfrac{2}{3} + \dfrac{2\cdot 5}{3\cdot 5} + \dfrac{2\cdot 5\cdot 8}{3\cdot 5\cdot 7} + \dfrac{2\cdot 5\cdot 8\cdot 11}{3\cdot 5\cdot 7\cdot 9}+ \cdots
$$
Check your answer
The series is absolutely convergent.
Video solution
The Root Test¶
The idea of the root test is similar to that of the ratio test, except that we take the limit of the $n^\text{th}$-root of the absolute series' terms.
Theorem¶
If $\lim\limits_{n\to\infty} \sqrt[n]{\lvert a_n \rvert} = L < 1$, then the series $\displaystyle\sum_{n=1}^{\infty} a_n$ is absolutely convergent;
If $\lim\limits_{n\to\infty} \sqrt[n]{\lvert a_n \rvert} = L > 1$ or $\lim\limits_{n\to\infty} \sqrt[n]{\lvert a_n \rvert } = \infty$ then the series $\displaystyle\sum_{n=1}^{\infty} a_n$ is divergent;
If $\lim\limits_{n\to\infty} \sqrt[n]{\lvert a_n \rvert} = 1$, then the Root Test is inconclusive; that is, no conclusion can be drawn about the convergence or divergence of $\sum a_n$.
Note. Just like with the Root Test, if $\lim\limits_{n\to\infty} \sqrt[n]{\lvert a_n \rvert} = 1$, by part (3) of the Root Test, we can not determine if the series is convergent or divergent. If you use the Root Test, and get $\lim\limits_{n\to\infty} \sqrt[n]{\lvert a_n \rvert} = 1$, DO NOT try to use the Ratio Test because the limit will also be 1, and vice versa.
9. Example¶
Use the Root Test to determine whether the series is convergent or divergent.
$$
\sum_{n=1}^{\infty}\dfrac{(-2)^{n}}{n^{n}}
$$
Check your answer
The series is absolutely convergent.
Video solution
10. Example¶
Use the Root Test to determine whether the series is convergent or divergent.
$$
\sum_{n=2}^{\infty}\left(\dfrac{n}{\ln(n)}\right)^{n}
$$
Check your answer
The series is divergent.
Video solution
Rearranging Sums¶
The Commutative Property of Addition states that if we rearrange the terms in a finite sum, the value of the sum remains unchanged; e.g., $a + b = b + a$.
So we must ask ourselves,
Do infinite sums behave like finite sums?
That is, if we rearrange the terms in an infinite sum will the value of sum remain unchanged?
It turns out that if $\sum a_n$ is an absolutely convergent series with sum $S$, then any rearrangement of $\sum a_n$ has the same sum $S$.
The same is not true of conditionally convergent series.
11. Example¶
Consider the absolutely convergent series
$$
\sum_{n=0}^{\infty} \dfrac{1}{2^n} = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots = 2
$$
If instead we decided to add the terms in a different order, we would still get the same value for the sum
$$
\sum_{n=0}^{\infty} \dfrac{1}{2^n} = \dfrac{1}{2} + 1 + \dfrac{1}{8} + \dfrac{1}{4} \cdots = 2
$$
Again, this is NOT the case if the series is conditionally convergent.
Any conditionally convergent series can be rearranged to give a different sum. Moreover, if $\sum a_n$ is a conditionally convergent series and $r$ is any real number, then there is a rearrangement of $\sum a_n$ that has a sum equal to $r$.
In other words, for every conditionally convergent series, there is a rearrangement of the terms for which the sum is equal to any real number of our choosing.
12. Example¶
Consider the alternating harmonic series, which we know is conditionally convergent. Later in the semester, we will show that
$$
\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} = 1 - \dfrac{1}{2} + \dfrac{1}{3} + \cdots = \ln 2.
$$
Multiplying this series by $\dfrac{1}{2}$, we get
$$
\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} = \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{6} + \cdots = \dfrac{1}{2}\ln 2.
$$
Since adding zeros does not affect the sum of the series, we can write the sum as
$$
\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n} = 0 + \dfrac{1}{2} + 0 - \dfrac{1}{4} + 0 + \dfrac{1}{6} + \cdots = \dfrac{1}{2}\ln 2.
$$
Notice that if we add both of these sums term-by-term we get
$$
1 + \dfrac{1}{3} - \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{7} - \dfrac{1}{4}... = \dfrac{3}{2}\ln 2.
$$
This sum is simply a rearrangement of the series we started with—the alternating harmonic series. However, the sums of these series are different.
Note. This argument does not rely on the fact that the exact sum of the alternating harmonic series is $\ln 2$. You could replace $\ln 2$ by $S$ throughout this argument and you will still end up with two different values to the "same" series.
13. Example¶
For which positive integers $k$ is the following series convergent?
$$
\sum_{n=1}^{\infty}\dfrac{(n!)^{2}}{(kn)!}
$$
Check your answer
Video solution
Discussion¶
Questions? You can ask them here.