Alternating series

Alternating series

  • Definition. An alternating series is a series whose terms are alternately positive and negative.
    • Examples: \[\begin{gather*} 1-\frac12+\frac13-\frac14+\frac15-\frac16+\dotsb=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\\ -\frac12+\frac23-\frac34+\frac45-\frac56+\frac67-\dotsb=\sum_{n=1}^\infty(-1)^n\frac{n}{n+1}. \end{gather*}\]
    • In general, \(\sum a_n\) is an alternating series, if for every \(n\), we have \[ a_n=(-1)^{n-1}b_n\text{ or }a_n=(-1)^nb_n \] where \(b_n>0\) for all \(n\). Note that we have \(b_n=|a_n|\).

Alternating series test

  • Alternating series test.
    • Let \(\sum a_n\) be an alternating series, and let \(b_n=|a_n|\).
    • Suppose that the sequence \(\{b_n\}\) is decreasing.
    • Suppose also that we have \(\lim_{n\to\infty}b_n=0\).
    • Then the series \(\sum a_n\) is convergent.
  • Example. Consider the series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\).
    • Since \(\frac{1}{n}\ge\frac{1}{n+1}\) for all \(n\ge1\), the sequence \(\{b_n\}=\left\{\frac{1}{n}\right\}\) is decreasing.
    • We have \(\lim_{n\to\infty}\frac{1}{n}=0\).
    • Therefore, the series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\) is convergent.
  • Exercises. 11.5: 3, 5, 11, 13, 17, 19

Alternating series estimation theorem

  • Alternating series estimation theorem. Suppose that \(s=\sum(-1)^nb_n\), where \(\{b_n\}\) is a sequence that is positive, decreasing, and converges to 0.
    • Then for any \(n\), we have \[ |R_n|=|s-s_n|\le b_{n+1}. \]
  • Example. Consider the series \(\sum_{n=0}^\infty(-1)^n\frac{2^n}{3\cdot 5^n}\).
    • You can check that is satisfies the conditions of the Alternating series test.
    • Let's find an approximation of \(s=\sum_{n=0}^\infty(-1)^n\frac{2^n}{3\cdot 5^n}\) correct to 3 decimal places.
    • For this, we need to find an \(n\) such that \(\frac{2^n}{3\cdot 5^n}<0.0005\).
    • For \(n=8\), we have \(\frac{2^8}{3\cdot 5^8}\approx0.0002\).
    • This means that \(s_8\approx0.5554\) is correct up to 3 decimal places.
    • Click here for the minimalistic SAGE code I used to perform the approximation.
  • Exercises. 11.5: 23, 27.

Absolute convergence. The Ratio and Root tests.

Absolute convergence

  • Definition. Let \(\sum a_n\) be a series.
    • Suppose that the series \(\sum|a_n|\) is convergent.
    • Then we say that the series \(\sum a_n\) is absolutely convergent.
  • Examples.
    • The series \[ \sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n^3}=1-\frac18+\frac{1}{27}-\frac{1}{64}+\dotsb \] is absolutely convergent, because the series \[ \sum_{n=1}^\infty\frac{1}{n^3} \] is convergent as \(3>1\).
    • We have seen that by the Alternating series test, the alternating harmonic series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\) is convergent. But it's not absolutely convergent, because the harmonic series \(\sum_{n=1}^\infty\frac{1}{n}\) is divergent.

Conditional convergence

  • Definition. If a series is convergent but not absolutely convergent, we say it is conditionally convergent.
  • Theorem. If a series is absolutely convergent, then it is convergent.
  • Example. Consider the series \(\sum_{n=1}^\infty\frac{\sin n}{n^2}\).
    • Since we have \[ 0\le\frac{|\sin n|}{n^2}\le\frac{1}{n^2}, \] the Comparison theorem implies that \(\sum_{n=1}^\infty\frac{\sin n}{n^2}\) is absolutely convergent, and therefore it is convergent.

The Ratio test

  • Ratio test. Let \(\sum a_n\) be a series.
    • If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L<1\), then \(\sum a_n\) is absolutely convergent.
    • If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L>1\) or \(\infty\), then \(\sum a_n\) is divergent.
    • If \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=1\) or divergent, then the Ratio test is inconclusive.
  • Example. Consider the series \(\sum_{n=1}^\infty\frac{n^2}{3^n}\).
    • We have \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^2}{n^2}\cdot\frac{3^n}{3^{n+1}}\to\frac13<1\), therefore the series is convergent.
  • Example. Let's revisit the series \(\sum_{n=1}^\infty\frac{n^n}{n!}\).
    • We have \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}\)==()n=(1+)ne>1$, therefore the series is divergent.
  • Exercises. 11.6: 7, 9, 17, 21.

The Root test

  • Root test. Let \(\sum a_n\) be a series.
    • If \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=L<1\), then \(\sum a_n\) is absolutely convergent.
    • If \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=L>1\) or \(\infty\), then \(\sum a_n\) is divergent.
    • If \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=1\) or divergent, then the Root test is inconclusive.
    • Note that the Ratio and Root tests yield a limit of 1 at the same time. That is, if the limit is 1 for one of the tests, then you shouldn't try the other test, but some other method.
  • Example. Consider the series \(\sum_{n=1}^\infty\left(\frac{n-1}{n^2+2n+8}\right)^{2n}\).
    • Since we have \(\sqrt[n]{a_n}=\left(\frac{n-1}{n^2+2n+8}\right)^2\to0\), the series is absolutely convergent.
  • Exercises. 11.6: 27, 29.
  • More exercises. 11.6: 31, 33, 35, 37.


  • Now let's look at something very weird.
  • Let \(\sum a_n\) be a series. A rearrangement of \(\sum a_n\) is a series \(\sum b_n\), with exactly the same terms, but in a different order. For example, we could have \[ \sum b_n=a_2+a_4+a_1+a_5+a_3+\dotsb \]
    • Theorem. If \(\sum a_n\) is absolutely convergent, then any rearrangement of it will have the same sum.
  • It turns out that this is not true if \(\sum a_n\) is conditionally convergent!
    • Theorem. Let \(\sum a_n\) be a conditionally convergent series, and \(r\) any real number. Then there exists a rearrangement of \(\sum a_n\) with sum \(r\).
    • If you're interested, you can find a proof sketch in 11.6.52.


  • Example. Consider the alternating harmonic series \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\).
    • We have \[\begin{equation}\tag{I} 1-\frac12+\frac13-\frac14+\frac15-\dotsb=\ln2. \end{equation}\]
    • Multiplying by \(\frac12\), we get \[ \frac12-\frac14+\frac16-\frac18+\frac{1}{10}-\dotsb=\frac12\ln2. \]
    • We can insert zeros in this to get \[\begin{equation}\tag{II} 0+\frac12+0-\frac14+0+\frac16+\dotsb=\frac12\ln2. \end{equation}\]
    • Now (I)\(+\)(II) is \[ 1+\frac13-\frac12+\frac15+\frac17-\frac14+\dotsb=\frac32\ln2, \] which you can check is a rearrangement of \(\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\).