In the last section the harmonic series
\[\sum_{n=1}^{\infty} \,\frac{1}{n} ~=~ 1 \;+\; \frac{1}{2} \;+\; \frac{1}{3} \;+\; \frac{1}{4} \;+\; \frac{1}{5} \;+\; \cdots \nonumber \]
was shown to diverge. If you were to alternate the signs of successive terms, as in
\[\label{eqn:altharmonic} \sum_{n=1}^{\infty} \,\frac{(-1)^{n-1}}{n} ~=~ 1 \;-\; \frac{1}{2} \;+\; \frac{1}{3} \;-\; \frac{1}{4} \;+\; \frac{1}{5} \;-\; \cdots \]
then it turns out that this new series—called an alternating series—converges, due to the following test:
The condition for the test means that \(\abs{a_{n+1}} \le \abs{a_n}\) for all \(n\) and \(\abs{a_n} \rightarrow 0\) as \(n \rightarrow \infty\). To see why the test works, consider the alternating series given above by formula ([eqn:altharmonic]), with \(a_n=\frac{-1^{n-1}}{n}\). The odd-numbered partial sums \(s_1\), \(s_3\), \(s_5\), \(\ldots\), can be written as
\[s_1 ~=~ 1 \quad,\quad s_3 ~=~ 1 \;-\; \underbrace{\left(\frac{1}{2} - \frac{1}{3}\right)}_{~>~ 0} \quad,\quad s_5 ~=~ 1 \;-\; \underbrace{\left(\frac{1}{2} - \frac{1}{3}\right)}_{~>~ 0} \;-\; \underbrace{\left(\frac{1}{4} - \frac{1}{5}\right)}_{~>~ 0} \quad,\quad\ldots \nonumber \]
while the even-numbered partial sums \(s_2\), \(s_4\), \(s_6\), \(\ldots\), can be written as
\[s_2 ~=~ 1 \;-\; \frac{1}{2} \quad,\quad s_4 ~=~ 1 \;-\; \frac{1}{2} \;+\; \underbrace{\left(\frac{1}{3} - \frac{1}{4}\right)}_{~>~ 0} \quad,\quad s_6 ~=~ 1 \;-\; \frac{1}{2} \;+\; \underbrace{\left(\frac{1}{3} - \frac{1}{4}\right)}_{~>~ 0} \;+\; \underbrace{\left(\frac{1}{5} - \frac{1}{6}\right)}_{~>~ 0} \quad,\quad\ldots \nonumber \]
Thus the odd-numbered partial sums \(\seq{s_{2n-1}}\) are decreasing from \(s_1=1\), and the even-numbered ones \(\seq{s_{2n}}\) are increasing from \(s_2=1-\frac{1}{2}=\frac{1}{2}\), with \(\frac{1}{2} < s_n < 1\) for all \(n\), i.e. the partial sums are bounded. So by the Monotone Bounded Test, both sequences \(\seq{s_{2n-1}}\) and \(\seq{s_{2n}}\) must converge. Since \(s_{2n} - s_{2n-1} = a_{2n} = \frac{-1}{2n}\) for all \(n \ge 1\), then
\[\lim_{n \to \infty} \,(s_{2n} - s_{2n-1}) ~=~ \lim_{n \to \infty} \;\frac{-1}{2n} ~=~ 0 \quad\Rightarrow\quad \lim_{n \to \infty} \;s_{2n} ~=~ \lim_{n \to \infty} \;s_{2n+1} \nonumber \]
Thus, the partial sums \(s_n\) have a common limit, so the series converges. Notice that the key to the convergence was having the terms decreasing in absolute value to zero.
Example \(\PageIndex{1}\)
Determine if \(~\displaystyle\sum_{n=2}^{\infty} \,\dfrac{(-1)^{n}}{\ln \,n}~\) is convergent.
Solution
For the general term \(a_n = \frac{(-1)^{n}}{\ln \,n}\), since \(\ln\,(n+1) > \ln\,n\) for \(n\ge 2\) and \(\ln\,n \rightarrow \infty\) as \(n \rightarrow \infty\), then \(\abs{a_n}\) decreases to 0 as \(n \rightarrow \infty\). Thus, by the Alternating Series Test the series converges.
The series \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) converges by the Alternating Series Test, though the series \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges. This makes \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) an example of a conditionally convergent series:
For example, \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) is not absolutely convergent, since \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges.
Example \(\PageIndex{1}\)
Is \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{(-1)^{n-1}}{n^2}~\) conditionally convergent or absolutely convergent?
Solution
Since \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges (by the p-series Test) then \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}\) is absolutely convergent.
It turns out that absolute convergence implies ordinary convergence:
The test is obvious if the terms \(a_n\) are all positive, so assume the series has both positive terms (denoted by the sequence \(\seq{a_{\text{pos}}}\)) and negative terms (denoted by \(\seq{a_{\text{neg}}}\)). Then the series can be decomposed into the difference of two series:
\[\sum a_n ~=~ \sum a_{\text{pos}} ~-~ \sum \;\abs{a_{\text{neg}}} \nonumber \]
Since each of the sums on the right side of the equation is part of the convergent series \(\sum \,\abs{a_n}\), then each sum itself converges (being part of a finite sum). Thus their difference, namely \(\sum a_n\), is finite, i.e. \(\sum a_n\) converges.
The test can be stated in the following logically equivalent manner:
One unusual feature of a conditionally convergent series is that its terms can be rearranged to converge to any number, a result known as Riemann’s Rearrangement Theorem. For example, the alternating harmonic series
\[1 \;-\; \frac{1}{2} \;+\; \frac{1}{3} \;-\; \frac{1}{4} \;+\; \frac{1}{5} \;-\; \cdots \nonumber \]
consists of one divergent series of positive terms subtracted from another series of positive terms, namely:
\[1 \;-\; \frac{1}{2} \;+\; \frac{1}{3} \;-\; \frac{1}{4} \;+\; \frac{1}{5} \;-\; \cdots ~=~ \left(1 \;+\; \frac{1}{3} \;+\; \frac{1}{5} \;+\; \frac{1}{7} \;+\; \cdots\right) ~-~ \left(\frac{1}{2} \;+\; \frac{1}{4} \;+\; \frac{1}{6} \;+\; \cdots\right) \nonumber \]
The idea is that since the first series on the right diverges then that series has some partial sum that could be made just larger than any positive number \(A\). Likewise, since the second series on the right that is being subtracted also diverges, it has some partial sum that when subtracted from the first partial sum results in a number just less than \(A\). Continue like this indefinitely, first adding a partial sum to get a number just bigger than \(A\) then subtracting another partial sum to get just less than \(A\). Since the terms in each series approach zero, the overall series can be made to converge to \(A\)! It also turns out that an absolutely convergent series does not have this feature—any rearrangement of terms results in the same sum.7
[sec9dot3]
For Exercises 1-5 determine whether the given alternating series is convergent. If convergent, then determine if it is conditionally convergent or absolutely convergent.
5
\(\bigsum{n = 1}{\infty}~ \dfrac{(-1)^{n - 1}}{\sqrt{n}}\)
\(\bigsum{n = 1}{\infty}~ \dfrac{(-1)^{n - 1}\;n\,!}{2^n}\)
\(\bigsum{n = 1}{\infty}~ \dfrac{(-1)^{n - 1}}{2n-1}\)
\(\bigsum{n = 2}{\infty}~ \dfrac{(-1)^{n}}{n\;\ln\,n}\)
\(\bigsum{n = 1}{\infty}~ \dfrac{(-1)^{n-1}}{n\,!}\)
Rearrangements of a divergent alternating series can make it appear to converge to different numbers. For example, find different rearrangements of the terms in the divergent series
\[\sum_{n=0}^{\infty} \;(-1)^n ~=~ 1 \;-\; 1 \;+\; 1 \;-\; 1 \;+\; \cdots \nonumber \]
so that the series appears to converge to 0, 1, -1, and 2.
A guest arrives at the Aleph Null Hotel,8 which has an infinite number of rooms, numbered Room 0, Room 1, Room 2, and so on. The hotel manager says all the rooms are taken, but he can still give the guest his own room. How is that possible? Would it still be possible if an infinite number of guests showed up, each wanting his own room? Explain your answers.
