increasing decreasing
Alternating series are series whose terms alternate in sign between positive
and negative. There is a powerful convergence test for alternating series.
Many of the series convergence tests that have been introduced so far are stated with the assumption that all terms in the series are nonnegative. Indeed, this condition is assumed in the Integral Test, Ratio Test, Root Test, Comparison Test and Limit Comparison Test. In this section, we study series whose terms are not assumed to be strictly positive. In particular, we are interested in series whose terms alternate between positive and negative (aptly named alternating series). It turns out that there is a powerful test for determining that a series of this form converges.
Let \(\{a_n\}_{n=1}^\infty \) be a sequence of positive numbers. An alternating series is a series of the form
\[ \sum _{n=1}^\infty (-1)^n a_n = -a_1 + a_2 - a_3 + a_4 - \cdots \]
or of the form
\[ \sum _{n=1}^\infty (-1)^{n+1}a_n = a_1 - a_2 + a_3 - a_4 + \cdots . \]
As usual, this definition can be modified to include series whose indexing starts somewhere other than \(n=1\).
The geometric series
\[ \sum _{n=0}^\infty \left (-\frac {2}{3}\right )^n = \sum _{n=0}^\infty \left (-1\right )^n \left (\frac {2}{3}\right )^n = 1 - \frac {2}{3} + \frac {4}{9} - \frac {8}{27} + \cdots \]
is alternating. In general, a geometric series with ratio \(r < 0\) is alternating, since
\[ \sum _{n=0}^\infty a r^n = \sum _{n=0}^\infty (-1)^n a |r|^n. \]
Recall the harmonic series
\[ \sum _{n=1}^\infty \frac {1}{n} = 1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} + \cdots , \]
which is perhaps the simplest example of a divergent series whose terms approach
zero as \(n\) approaches \(\infty \). A similarly important example is the alternating harmonic
series
\[ \sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} = 1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4} + \cdots . \]
The terms of this series, of course, still approach zero, and their absolute values are
monotone decreasing. Because the series is alternating, it turns out that this is
enough to guarantee that it converges. This is formalized in the following theorem.
Alternating Series Test Let \(\{a_n\}\) be a sequence whose terms are eventually positive and
nonincreasing and \(\lim _{n\to \infty }a_n=0\). Then, the series
\[ \sum _{n=1}^\infty (-1)^{n}a_n \qquad \text {and}\qquad \sum _{n=1}^\infty (-1)^{n+1}a_n \]
both converge. Compared to our convergence tests for series with strictly positive terms, this test is strikingly simple. Let us examine why it might be true by considering the partial sums of the alternating harmonic series. The first few partial sums with odd index are given by \begin{align*} s_1 &= 1 \\ s_3 &= 1 - \frac {1}{2} + \frac {1}{3} = s_1 - \left (\frac {1}{2} - \frac {1}{3}\right ) = \frac {5}{6} \\ s_5 &= s_3 - \frac {1}{4} + \frac {1}{5} = s_3 - \left ( \frac {1}{4} - \frac {1}{5}\right ) = \frac {47}{60}. \end{align*}
Note that a general odd partial sum is of the form
\[ s_{2n+1} = s_{2n-1} - \left (\frac {1}{2n} - \frac {1}{2n+1}\right ), \]
and the quantity in the parentheses is positive. We conclude that:
Moreover, the sequence of odd partial sums is bounded below by zero, since
\[ s_{2n+1} = \left (1 - \frac {1}{2}\right ) + \left (\frac {1}{3} - \frac {1}{4}\right ) + \cdots + \left (\frac {1}{2n-1}-\frac {1}{2n}\right ) + \frac {1}{2n+1}, \]
and each quantity in parentheses is positive.
We conclude that the sequence \(\{s_1,s_3,s_5,\ldots \}\) of odd partial must converge to a finite limit by
applying
The Fundamental Theorem of Calculus. The Monotone Convergence
Theorem. The Ratio Test.
Next consider the sequence \begin{align*} s_2 &= 1 - \frac {1}{2} = \frac {1}{2} \\ s_4 &= 1- \frac {1}{2} + \frac {1}{3} - \frac {1}{4} = \frac {7}{12} \\ s_6 &= 1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4} + \frac {1}{5} - \frac {1}{6} = \frac {37}{60} \\ &\vdots \end{align*}
of even partial sums.
The sequence \(\{s_2,s_4,s_6,\ldots \}\) of even partial sums defined above is
increasing and bounded, and
therefore converges by the Monotone Convergence Theorem. decreasing and
bounded, and therefore converges by the Monotone Convergence Theorem.
Finally, we use
\[ \lim _{n \rightarrow \infty } \left (s_{2n+1} - s_{2n} \right ) = \lim _{n \rightarrow \infty } \frac {1}{2n+1} = 0 \]
and the fact that the limits of the sequences \(\{s_{2n+1}\}\) and \(\{s_{2n}\}\) are finite to conclude that
\[ \lim _{n \rightarrow \infty } s_{2n+1} = \lim _{n \rightarrow \infty } s_{2n} = \lim _{n\rightarrow \infty } s_n. \]
It follows that the alternating harmonic series converges.
With slight modification, the argument given above can be used to prove that the Alternating Series Test holds in general.
Does the series \(\sum _{n=1}^\infty (-1)^n\frac {n}{n^2+1}\) converge?
The terms of the sequence \(\left \{n/(n^2+1)\right \}\) are positive and nonincreasing, so we can apply the Alternating Series Test. Since
\[ \lim _{n \rightarrow \infty } \frac {n}{n^2+1} = 0, \]
the Alternating Series Test implies that the series converges. Does the alternating series test apply to the series
2025-01-06 18:39:45 \[ \sum _{n=1}^\infty (-1)^{n+1}\frac {|\sin n|}{n^2}? \]
yes no
The underlying
sequence is \(\left (a_n\right ) = |\sin n|/n\). 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 \(\left (a_n\right )\) 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.