The various concepts associated with sequences and series are reviewed.
All of the series convergence tests we have used require that the underlying sequence \(\seq {a_n}\)
be a positive sequence. We can actually relax this and state that there must be an \(N>0\)
such that \(a_n>0\) for all \(n>N\); that is, \(\seq {a_n}\) is positive for all but a finite number of values of \(n\). We’ve
also stated this by saying that the tail of the sequence must have positive
terms. In this section we explore series whose summation includes negative
terms.
1 Alternating series test
We start with a very specific form of series, where the terms of the summation
alternate between being positive and negative.
Let
\(\seq {a_n}\) be a positive sequence. An
alternating series is a series of either the form
\[ \sum _{n=1}^\infty (-1)^na_n\qquad \text {or}\qquad \sum _{n=1}^\infty (-1)^{n+1}a_n. \]
In essence, the signs of the terms of \(\seq {a_n}\) alternate between positive and negative.
Recall that the terms of the harmonic series come from the harmonic sequence
\(\seq {a_n} = \seq {1/n}\). An important alternating series is the alternating harmonic series:
\[ \sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} = 1-\frac 12+\frac 13-\frac 14+\frac 15-\frac 16+\cdots \]
Geometric series are also alternating series when \(r<0\). For instance, if \(r=-1/2\), the geometric
series is
\[ \sum _{n=0}^\infty \left (\frac {-1}{2}\right )^n = 1-\frac 12+\frac 14-\frac 18+\frac 1{16}-\frac 1{32}+\cdots \]
We know that geometric series converge when \(|r|<1\) and have the sum:
\[ \sum _{n=0}^\infty r^n = \frac 1{1-r}. \]
When \(r=-1/2\) as above,
we find
\[ \sum _{n=0}^\infty \left (\frac {-1}{2}\right )^n = \frac 1{1-(-1/2)} = \frac 1{3/2} = \frac 23. \]
A powerful convergence theorem exists for other alternating series that meet a few
conditions.
Alternating Series Test Let
\(\seq {a_n}\) be a positive, nonincreasing sequence where
\(\lim _{n\to \infty }a_n=0\). Then
\[ \sum _{n=1}^\infty (-1)^{n}a_n \qquad \text {and}\qquad \sum _{n=1}^\infty (-1)^{n+1}a_n \]
converge.
Does the alternating series test apply to the series
\[ \sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} ? \]
yes no
This is the
alternating harmonic series as seen previously. The underlying sequence is \(\seq {a_n} = \seq {1/n}\), which is
positive, decreasing, and approaches 0 as \(n\to \infty \). Therefore we can apply the alternating
series test and conclude this series converges.
Does the alternating series test apply to the series
\[ \sum _{n=1}^\infty \frac {(-1)^n\ln n}{n}? \]
yes no
The underlying
sequence is
\(\seq {a_n} = \seq {\ln n/n}\). This is positive and approaches
\(0\) as
\(n\to \infty \) (use L’Hôpital’s Rule). However, the
sequence is not decreasing for all
\(n\). It is straightforward to compute
\(a_1=0\),
\(a_2\approx 0.347\),
\(a_3\approx 0.366\), and
\(a_4\approx 0.347\): the
sequence is increasing for at least the first
\(3\) terms.
However, we do not “give-up” and immediately conclude that we cannot apply the
alternating series test. Rather, consider the long term behavior of \(\seq {a_n}\). Treating \(a_n=a(n)\) as a
continuous function of \(n\) defined on \((1,\infty )\), we can take its derivative:
\[ a'(n) = \frac {1-\ln n}{n^2}. \]
The derivative
is negative for all
\(n\geq 3\) (actually, for all
\(n>e\)), meaning
\(a(n)=a_n\) is decreasing on
\((3,\infty )\). We can
now apply the alternating series test to the series when we start with
\(n=3\) and
conclude that
\(\sum _{n=3}^\infty (-1)^n\frac {\ln n}{n}\) converges; adding the terms with
\(n=1\) and
\(n=2\) do not change the
convergence.
The important lesson here is that as before, if a series fails to meet the criteria of the
alternating series test on only a finite number of terms, we can still apply the test.
Does the alternating series test apply to the series
\[ \sum _{n=1}^\infty (-1)^{n+1}\frac {|\sin n|}{n^2}? \]
yes no
The underlying
sequence is
\(\seq {a_n} = |\sin n|/n\). This sequence is positive and approaches
\(0\) as
\(n\to \infty \). However, it is not a
decreasing sequence; the value of
\(|\sin n|\) oscillates between
\(0\) and
\(1\) as
\(n\to \infty \). We cannot remove a
finite number of terms to make
\(\seq {a_n}\) decreasing, therefore we cannot apply the alternating
series test.
Keep in mind that this does not mean we conclude the series diverges; in fact, it does
converge. We are just unable to conclude this based on the alternating series test.
2 Approximating alternating series
While there are many factors involved when studying rates of convergence, the
alternating structure of an alternating series gives us a powerful tool when
approximating the sum of a convergent series.
Alternating Series Approximation Let
\(\seq {a_n}\) be a sequence that satisfies the hypotheses of
the alternating series test, let
\(S_n\) be the
\(n\)th partial sum, and let
\[ L = \sum _{n=1}^\infty (-1)^{n}a_n\qquad \text {or}\qquad L=\sum _{n=1}^\infty (-1)^{n+1}a_n. \]
Then
- \(|L-S_n| \leq a_{n+1}\), and
- \(L\) is between \(S_n\) and \(S_{n+1}\).
In this case, \(R_n=L-S_n\) is called the \(n\)th remainder of the series.
Here is the basic idea behind this theorem. Say we have an alternating sequence, \(\seq {a_n}\).
Let’s assume the first term is positive, so the second is negative, and so on. We add
the first two numbers and get some number \(S_2\). Now \(S_2\) is smaller than \(S_1 = a_1\), because we
subtracted something from \(a_1\). Next, we add on the third term, \(a_3\), to get the partial sum \(S_3\).
This \(S_3\) is bigger than \(S_2\), because the sequence is alternating, but is smaller than \(S_1\),
because the sequence is decreasing.
If we know the series converges to some \(L\), we can see that we must be bouncing back
and forth around \(L\) as we add and subtract terms. At one point, \(S_n\) is larger than \(L\), and
then subtracting off the next term makes the partial sum smaller than \(L\). In other
words, the true limit \(L\) must be between \(S_n\) and \(S_{n+1}\). Imagine plotting \(\seq {a_n}\), \(\seq {S_n}\), and \(L\) on a number
line. (Or, try it yourself with the alternating harmonic series!) can be no
further from \(S_n\) than whatever the next term in the sequence is. How do we
get from \(S_n\) to \(S_{n+1}\)? By adding (or subtracting) \(a_{n+1}\), which takes us back “across”
\(L\) again. In other words, the distance between \(L\) and \(S_n\) can be no more than
\(a_{n+1}\).
See if you can use these same ideas to prove the alternating series test!
Let’s see an example of approximating an alternating series.
Approximate the sum of the alternating harmonic series with an error less than
\(10^{-2}\).
Look at the
\((n+1)\)th term.
\begin{align*} \frac {1}{n+1} &< 10^{-2}\\ n+1&> 100\\ n&> 99 \end{align*}
Using a computer, we can see that
\[ S_{99} = 0.69849462\dots \]
We will see later that the true value of this series
is
\(\ln (2)\). Comparing
\begin{align*} \ln (2) &= 0.69314718\dots \qquad \text {and}\\ S_{99} &= 0.69849462\dots \end{align*}
we have our desired accuracy.
3 Absolute convergence versus conditional convergence
It is an interesting result that the harmonic series,
\[ \sum _{n=1}^\infty \frac 1n \]
diverges, yet the alternating
harmonic series,
\[ \sum _{n=1}^\infty (-1)^{n+1}\frac 1n, \]
converges. The notion that simply alternating the signs of the terms
in a series can change a series from divergent to convergent leads us to the following
definitions.
- A series \(\sum _{n=1}^\infty a_n\) converges absolutely if \(\sum _{n=1}^\infty |a_n|\) converges.
- A series \(\sum _{n=1}^\infty a_n\) converges conditionally if \(\sum _{n=1}^\infty a_n\) converges but \(\sum _{n=1}^\infty |a_n|\) diverges.
Note, in the definition above, \(\sum _{n=1}^\infty a_n\) is not necessarily an alternating series; it may just
have some negative terms.
Does the series
\[ \sum _{n=1}^\infty (-1)^n\frac {n+3}{n^2+2n+5} \]
converge absolutely, converge conditionally, or diverge?
The
series converges conditionally. The series converges absolutely. The series
diverges.
We can show the series
\[ \sum _{n=1}^\infty \left |(-1)^n\frac {n+3}{n^2+2n+5}\right |= \sum _{n=1}^\infty \frac {n+3}{n^2+2n+5} \]
diverges using the limit comparison test,
comparing with
\(1/n\). The series
\[ \sum _{n=1}^\infty (-1)^n\frac {n+3}{n^2+2n+5} \]
converges using the alternating series test; we conclude
the series converges conditionally.
Does the series
\[ \sum _{n=1}^\infty (-1)^n\frac {n^2+2n+5}{2^n} \]
converge absolutely, converge
conditionally, or diverge?
The series converges conditionally. The series
converges absolutely. The series diverges.
We can show the series
\[ \sum _{n=1}^\infty \left |(-1)^n\frac {n^2+2n+5}{2^n}\right |=\sum _{n=1}^\infty \frac {n^2+2n+5}{2^n} \]
converges using
the ratio test. Therefore we conclude
\[ \sum _{n=1}^\infty (-1)^n\frac {n^2+2n+5}{2^n} \]
converges absolutely.
Does the series
\[ \sum _{n=3}^\infty (-1)^n\frac {3n-3}{5n-10} \]
converge absolutely, converge conditionally, or diverge?
The series converges
conditionally. The series converges absolutely. The series diverges.
The
series
\[ \sum _{n=3}^\infty \left |(-1)^n\frac {3n-3}{5n-10}\right | = \sum _{n=3}^\infty \frac {3n-3}{5n-10} \]
diverges using the divergence test, so it does not converge absolutely.
The series
\[ \sum _{n=3}^\infty (-1)^n\frac {3n-3}{5n-10} \]
fails the conditions of the alternating series test as
\((3n-3)/(5n-10)\) does not
approach
\(0\) as
\(n\to \infty \). We can state further that this series diverges; as
\(n\to \infty \), the series
effectively adds and subtracts
\(3/5\) over and over. This causes the sequence of partial
sums to oscillate and not converge. Therefore the series diverges.
Knowing that a series converges absolutely allows us to make two important
statements. The first, given in the following theorem, is that absolute convergence is
“stronger” than regular convergence. That is, just because
\[ \sum _{n=1}^\infty a_n \]
converges, we cannot
conclude that
\[ \sum _{n=1}^\infty |a_n| \]
will converge, but knowing a series converges absolutely tells us that \(\sum _{n=1}^\infty a_n\)
will converge.
One reason this is important is that our convergence tests all require that the
underlying sequence of terms be positive. By taking the absolute value of the terms
of a series where not all terms are positive, we are often able to apply an appropriate
test and determine absolute convergence. This, in turn, determines that the series we
are given also converges.
The second statement relates to rearrangements of series. When dealing with
a finite set of numbers, the sum of the numbers does not depend on the
order in which they are added. (So \(1+2+3 = 3+1+2\).) One may be surprised to find out that
when dealing with an infinite set of numbers, the same statement does not
always hold true: some infinite lists of numbers may be rearranged in different
orders to achieve different sums. The theorem states that the terms of an
absolutely convergent series can be rearranged in any way without affecting the
sum.
Absolute Convergence Let
\(\sum _{n=1}^\infty a_n\) be a series that converges absolutely. Let
\(\seq {b_n}\) be any
rearrangement of the sequence
\(\seq {a_n}\). Then
\[ \sum _{n=1}^\infty b_n = \sum _{n=1}^\infty a_n. \]
This theorem states that rearranging the terms of an absolutely convergent series
does not affect its sum. Making such a statement implies that perhaps the sum of a
conditionally convergent series can change based on the arrangement of terms.
Indeed, it can. The Riemann rearrangement theorem (named after Bernhard
Riemann) states that any conditionally convergent series can have its terms
rearranged so that the sum is any desired value, including \(\infty \)!
As an example, consider the alternating harmonic series once more. We have stated
that
\[ \sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} =1-\frac 12+\frac 13-\frac 14+\frac 15-\frac 16+\frac 17\cdots = \ln 2. \]
Consider the rearrangement where every positive term is followed by two
negative terms:
\[ 1-\frac 12-\frac 14+\frac 13-\frac 16-\frac 18+\frac 15-\frac 1{10}-\frac 1{12}\cdots \]
(Convince yourself that these are exactly the same numbers as
appear in the alternating harmonic series, just in a different order.) Now group some
terms and simplify:
\begin{align*} \left (1-\frac 12\right )-\frac 14+\left (\frac 13-\frac 16\right )-\frac 18+\left (\frac 15-\frac 1{10}\right )-\frac 1{12}+\cdots &= \\ \frac 12-\frac 14+\frac 16-\frac 18+\frac 1{10}-\frac {1}{12}+\cdots &= \\ \frac 12\left (1-\frac 12+\frac 13-\frac 14+\frac 15-\frac 16+\cdots \right ) & = \frac 12\ln 2. \end{align*}
By rearranging the terms of the series, we have arrived at a different sum!
One could try to argue that the alternating harmonic series does not
actually converge to \(\ln 2\), and here is an example of such an argument.
According to the alternating series test, we know that this series
converges to some number \(L\). If, as our intuition tells us should be true,
the rearrangement does not change the sum, then we have just seen
that \(L = L/2\). The only possibility for \(L\) is then \(L=0\). But the alternating series
approximation theorem quickly shows that \(L>0\). The only conclusion is
that the rearrangement did, contrary to our intuition, change the
sum.
The fact that conditionally convergent series can be rearranged to equal any number
is really an incredible result.
While series are worthy of study in and of themselves, our ultimate goal within
calculus is the study of power series, which we will consider in the next section. We
will use power series to create functions where the output is the result of an infinite
summation.