Calculus II Home

## Absolute and Conditional Convergence

An alternating series is a series of the form $\sum_{n=1}^{\infty} (-1)^n a_n=-a_1+a_2-a_3+a_4-\cdots \text{ or } \sum_{n=1}^{\infty} (-1)^{n-1} a_n=a_1-a_2+a_3-a_4+\cdots,$ where $$\{a_n\}$$ is a sequence of positive real numbers.

Example. The following alternating series is called the alternating harmonic series $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots.$

Alternating Series Test: The alternating series $$\displaystyle\sum_{n=1}^{\infty} (-1)^n a_n$$ and $$\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n$$ are convergent if $$\{a_n\}$$ is a positive and decreasing sequence that converges to $$0$$, i.e.,

1. $$a_n>0$$ for all integers $$n\geq 1$$,

2. $$a_{n+1} < a_n$$ for all integers $$n\geq 1$$, and

3. $$\displaystyle\lim_{n\to \infty} a_n=0$$.

Consider the alternating series $$\displaystyle\sum_{n=1}^{\infty} (-1)^n a_n$$ where $$\{a_n\}$$ is a positive and decreasing sequence that converges to $$0$$. Since $$\{a_n\}$$ is positive and decreasing, $s_{2n}=s_{2n-2}+(-a_{2n-1}+a_{2n}) < s_{2n-2}.$ Thus $$\{s_{2n}\}$$ is a decreasing sequence. Note that $s_{2n}=-a_1+(a_2-a_3)+\cdots+(a_{2n-2}-a_{2n-1})+a_{2n}.$ Since $$(a_2-a_3),\ldots,(a_{2n-2}-a_{2n-1}),a_{2n}$$ are positive, $$s_{2n} > -a_1$$. Thus $$\{s_{2n}\}$$ is a decreasing sequence that is bounded below. Then by the Monotone Convergence Theorem, $$\{s_{2n}\}$$ is convergent. Suppose $$\displaystyle\lim_{n\to \infty} s_{2n}=s$$. Since $$s_{2n+1}=s_{2n}-a_{2n+1}$$, $\lim_{n\to \infty} s_{2n+1}=\lim_{n\to \infty} s_{2n}- \lim_{n\to \infty} a_{2n+1}=s-0=s.$ Since both $$\{s_{2n}\}$$ and $$\{s_{2n+1}\}$$ converge to $$s$$, so does $$\{s_{n}\}$$. Therefore $\sum_{n=1}^{\infty} (-1)^n a_n=\lim_{n\to \infty} s_{n}=s.$

Example. Use the Alternating Series Test to show that $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$ is convergent.
Solution. The given series is $$\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1} a_n$$ where $$a_n=\frac{1}{n}$$. $$\{a_n\}=\{\frac{1}{n}\}$$ is a positive and decreasing sequence that converges to $$0$$ because

1. $$a_n=\frac{1}{n} > 0$$ for all integers $$n\geq 1$$,

2. $$a_{n+1}=\frac{1}{n+1} < \frac{1}{n}=a_n$$ for all integers $$n\geq 1$$, and
3. $$\displaystyle\lim_{n\to \infty} a_n =\lim_{n\to \infty}\frac{1}{n}=0$$.

Thus by the Alternating Series Test, $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$ is convergent.

Note:

1. To show that $$\{a_n\}$$ is decreasing where the formula of $$a_n=f(n)$$ is more complicated, show that $$f'(x) < 0$$ on $$(1,\infty)$$. For the above problem, $$f(x)=\frac{1}{x}$$ and $$f'(x)=-\frac{1}{x^2} < 0$$ on $$(1,\infty)$$. Then $$\{a_n\}=\{\frac{1}{n}\}$$ is decreasing.

2. In the above example, $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$ is convergent but $$\displaystyle \sum_{n=1}^{\infty} \left\vert \frac{(-1)^{n-1}}{n}\right\vert =\sum_{n=1}^{\infty} \frac{1}{n}$$ is divergent.

Definition. A series $$\displaystyle \sum_{n=1}^{\infty} a_n$$ is absolutely convergent if $$\displaystyle \sum_{n=1}^{\infty} |a_n|$$ is convergent. A series $$\displaystyle \sum_{n=1}^{\infty} a_n$$ is conditionally convergent if it is convergent but $$\displaystyle \sum_{n=1}^{\infty} |a_n|$$ is divergent (i.e., a conditionally convergent series is a convergent series that is not absolutely convergent).

Example. Recall that $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$ is convergent but $$\displaystyle \sum_{n=1}^{\infty} \left\vert \frac{(-1)^{n-1}}{n}\right\vert =\sum_{n=1}^{\infty} \frac{1}{n}$$ is divergent. Therefore $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$ is conditionally convergent but not absolutely convergent.

Theorem. An absolutely convergent series is convergent, i.e., if $$\displaystyle \sum_{n=1}^{\infty} |a_n|$$ is convergent, then $$\displaystyle \sum_{n=1}^{\infty} a_n$$ is also convergent.

Suppose that $$\displaystyle \sum_{n=1}^{\infty} |a_n|$$ is convergent. For all integers $$n\geq 1$$, $$-|a_n|\leq a_n \leq |a_n|$$. Adding $$|a_n|$$, we get $0\leq a_n+|a_n|\leq 2|a_n|.$ Since $$\displaystyle \sum_{n=1}^{\infty} |a_n|$$ is convergent, so is $$\displaystyle \sum_{n=1}^{\infty} 2|a_n|=2\sum_{n=1}^{\infty} |a_n|$$. By the Comparison Test, $$\displaystyle \sum_{n=1}^{\infty} (a_n+|a_n|)$$ is convergent. Finally $$\displaystyle \sum_{n=1}^{\infty} a_n$$ is also convergent as the difference of two convergent series: $\sum_{n=1}^{\infty} a_n= \sum_{n=1}^{\infty} (a_n+|a_n|) -\sum_{n=1}^{\infty} |a_n|.$

Example. Consider the series $$\displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n^2}$$ which has positive and negative terms. For all integers $$n\geq 1$$, $\left\vert\frac{\sin n}{n^2}\right\vert\leq \frac{1}{n^2}.$ Since $$\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$$ is convergent by the $$p$$-series Test with $$p=2 > 1$$, $$\displaystyle \sum_{n=1}^{\infty} \left\vert\frac{\sin n}{n^2}\right\vert$$ is convergent by the Comparison Test. Since $$\displaystyle \sum_{n=1}^{\infty} \left\vert\frac{\sin n}{n^2}\right\vert$$ is convergent, $$\displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n^2}$$ is absolutely convergent and consequently convergent.

To see the importance of the concept of absolute and conditional convergence of a series, we discuss rearrangements of the terms of a series. Consider the alternating harmonic series $$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$$ whose sum is obtained by evaluating the Taylor series of $$\ln(1+x)$$ at $$x=1$$. $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots =\ln 2.$ Now we rearrange the terms as follows. \begin{align*} &\;\;\left( 1-\frac{1}{2} \right)-\frac{1}{4}+ \left( \frac{1}{3}-\frac{1}{6} \right)-\frac{1}{8}+\cdots\\ &=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\cdots\\ &=\frac{1}{2}\left( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \right)\\ &=\frac{1}{2}\ln 2. \end{align*} We can even rearrange the terms to make it converge to any real number. This is due to the conditional convergence of the alternating harmonic series.

Riemann's Rearrangement Theorem: Let $$\displaystyle \sum_{n=1}^{\infty} a_n$$ be a conditionally convergent series and $$L$$ be a real number. Then there is a rearrangement of $$\displaystyle \sum_{n=1}^{\infty} a_n$$ that converges to $$L$$.

The above strange phenomenon does not happen for absolutely convergent series.
Theorem. Let $$\displaystyle \sum_{n=1}^{\infty} a_n$$ be an absolutely convergent series converging to $$L$$. Then any rearrangement of $$\displaystyle \sum_{n=1}^{\infty} a_n$$ also converges to $$L$$.

Last edited