Absolute & Conditional Convergence (College Board AP® Calculus BC): Revision Note

Absolute & conditional convergence

What is absolute convergence?

• Let $\sum _{n=1}^{\infty }{a}_n$ be an infinite series

  • If $\sum _{n=1}^{\infty }\left|a_n\right|$, converges, then $\sum _{n=1}^{\infty }{a}_n$ is said to converge absolutely

  • I.e. a series is absolutely convergent if the corresponding series of absolute values converges

• If a series converges absolutely then it converges

  • If you know that $\sum _{n=1}^{\infty }\left|a_n\right|$ converges, that tells you that $\sum _{n=1}^{\infty }{a}_n$ converges as well

    • There is no need to test $\sum _{n=1}^{\infty }{a}_n$ separately for convergence

• For example, consider $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{n^2}$

  • $\sum _{n=1}^{\infty }\frac{1}{n^2}$ converges which means $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{n^2}$ converges absolutely

What is conditional convergence?

• Let $\sum _{n=1}^{\infty }{b}_n$ be an alternating series (i.e. with alternating positive and negative terms)

  • If $\sum _{n=1}^{\infty }{b}_n$ converges, but $\sum _{n=1}^{\infty }\left|b_n\right|$ does not converge, then $\sum _{n=1}^{\infty }{b}_n$ is said to converge conditionally

  • I.e. a series is conditionally convergent if the series itself converges, but the corresponding series of absolute values does not

• For example, the alternating harmonic series $\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}$ converges conditionally

  • $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges

• Note that if $\sum _{n=1}^{\infty }\left|b_n\right|$ does not converge, that doesn't tell you anything about the convergence of $\sum _{n=1}^{\infty }{b}_n$

  • $\sum _{n=1}^{\infty }{b}_n$ may converge or diverge, and needs to be tested separately

Why is the difference between absolute and conditional convergence important?

• If a series converges absolutely, then any series obtained from it by regrouping or rearranging the terms in the series has the same value

• This is probably what you are used to when adding numbers

  • Rearranging the order or regrouping (by adding or removing brackets) doesn't change the sum

  • But this is only true when adding together finitely many numbers

    • It is not necessarily true for an infinite series!

• For example, consider the series $1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...$

  • This converges to a sum of $0$, as can be seen by writing it as

    $\begin{array}{l}=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...\\ =0+0+0+0+...\\ =0\end{array}$

  • It is conditionally convergent because the sum of absolute values gives

    $\begin{array}{l}=\left|1\right|+\left|-1\right|+\left|\frac{1}{2}\right|+\left|-\frac{1}{2}\right|+\left|\frac{1}{3}\right|+\left|-\frac{1}{3}\right|+\left|\frac{1}{4}\right|+\left|-\frac{1}{4}\right|+...\\ =1+1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}+...\\ =2\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...\right)\\ \to +\infty \end{array}$

    • That is 2 times the harmonic series, which we know diverges to $+\infty$

  • However the original series can be rearranged and regrouped to give a different sum

    $\begin{array}{l}=1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{6}-\frac{1}{3}+...\\ =1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{1}{2}\right)+\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{3}\right)+...\\ =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...\end{array}$

    • With that rearrangement it is equal to the alternating harmonic series, which converges to a sum of $\ln 2$

    • This appears to contradict it converging to $0$ above

• Conditionally convergent series can be regrouped and/or rearranged to give

  • a series with a different sum

  • or a series which does not converge to any sum