周次: 14 日期: May 25, 2022 节次: 1

In this lecture, we continue the discussion on the root test (for the series with non-negative terms). And we discuss alternating series, there’s a test called Leibniz’s or alternating series test. Alternating series converges in a manner that positive and negative terms cancelling each other by a large amount. With this fact, we define the absolute convergence and conditional convergence, and find them totally different from the rearrangement perspective.

The Root Test ( For the Series with Non-negative Terms)

The ratio test is convenient for the case when $a_{n+1}/a_n$ is relatively simple. For more complicated series, we may need other test method. For example

Guiding Example

Let $a_n=\{\begin{array}{lll}n/2^n,& n\text{ odd},1/2^n,& n\text{ even}\end{array}$, does $\sum _n{a}_n$ converge?

This is not a geometric series, and the $n$th term approaches $0$. So we do not know the series diverges. The Integral Test does not applicable as the sequence $a_n$ is not decreasing. The Ratio Test produces

$$\frac{a_{n+1}}{a_n}=\{\begin{array}{lll}\frac{1}{2n},& n\text{ odd}\frac{n+1}{2},& n\text{ even}\end{array}$$

As $n\to \mathrm{\infty }$, the ratio has no limit.

A test that will answer the question is the Root Test.

Theorem 13. The Root Test

Let $\sum a_n$ be a series with $a_n\ge 0$ for $n\ge N$, and suppose that

$$\underset{n}{lim}\sqrt[n]{a_n}=\rho .$$

Then

(a) the series converges if $\rho <1$

(b) the series diverges if $\rho >1$

(c) the test is inconclusive if $\rho =1.$

Proof.

Example 3. Applying the Root Test

(a) $\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{2^n}$

( Converges )

(b) $\sum _{n=1}^{\mathrm{\infty }}\frac{3^n}{(1+\frac{1}{n}{)}^{n^2}}$

( Diverges )

(c) $\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{1+n}\right)}^n$

( Converges )

Guiding Example (Revisited)

Let $a_n=\{\begin{array}{lll}n/2^n,& n\text{ odd},1/2^n,& n\text{ even}\end{array}$, does $\sum _n{a}_n$ converge?

( Converges )

Alternating Series

A series in which the terms are alternately positive and negative is an alternating series.

Examples (Alternating Series)

• $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{(-1{)}^{n+1}}{n}+\cdots$ (alternating harmonic series, converges)
• $-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots +\frac{(-1{)}^{n}4}{2^n}+\cdots$ (converges)
• $1-2+3-4+5-6+\cdots +(-1{)}^{n+1}n+\cdots$ (diverges)

These are alternating series.

Theorem 14. (The Alternating Series Test , Leibniz’s Theorem)

If $u_n\searrow ,0$ (starting from $n\ge N$), then the alternating series

$$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}{u}_n=u_1-u_2+u_3-u_4+\cdots$$

converges.

Proof.

Example 1. The alternating harmonic series

$$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{1}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \text{converges}$$

Graphical Interpretation on Convergence

Why for $u_n\searrow 0$, the series

$$u_1-u_2+u_3-u_4+u_5-\cdots$$

converges ?

From the graphical interpretation, we also learned that if the sum for alternating series is $L$, then the distance of the partial sum $s_n$ and the total sum $L$ satisfies

$$|L-s_n|<u_{n+1},\mathrm{\forall },n\ge N$$

Theorem 15. The Alternating Series Estimation Theorem

If the alternating series $\sum _n(-1{)}^{n+1}{u}_n$ satisfies $u_n\searrow 0$, then for $n\ge N$ the partial sum

$$s_n=u_1-u_2+u_3-\cdots +(-1{)}^{n+1}{u}_n$$

approximates the sum $L$ with an error whose absolute value is less than $u_{n+1},$ the numerical value of the first unused term. Furthermore, the remainder $L-s_n$ has the same sign as the first unused term.

Example 2. Try Theorem 15 on the series whose sum we know

$$\sum _n(-1{)}^n\frac{1}{2^n}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}|+\frac{1}{256}-\cdots$$

The sum of the series is $2/3$. And the partial sum indicating by the $|$ sign is $\frac{2}{3}(1-(-\frac{1}{2}{)}^6).$

The remainder, $\frac{2}{3}(-\frac{1}{2}{)}^8$ , is positive and less than $\frac{1}{2^8}.$

Absolute and Conditional Convergence

Definition. Absolutely Convergent

A series $\sum _n{a}_n$ converges absolutely (is absolutely convergent) if the corresponding series of absolute values, $\sum |a_n|$，converges.

Definition. Conditionally Convergent

A series that converges but does not converge absolutely converges conditionally.

Examples.

• All convergent series with non-negative terms converge absolutely
• $1+\frac{\sin (1)}{1^2}+\frac{\sin (2)}{2^2}+\frac{\sin (3)}{3^2}+\cdots$ converges absolutely
• $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ converges, but not absolutely. (Conditionally convergent)

Why Absolutely Convergent?

1. It is easier to test. (We have a number of tests for the series with non-negative terms)
2. Converges absolutely$⟹$ converges!

Theorem 16. The Absolute Convergence Test

If $\sum _n|a_n|$ converges, then $\sum _n{a}_n$ converges.

Proof.

Remark.

Absolutely convergent is a stronger than conditionally convergent. So a conditionally convergent series may fail to be an absolutely convergent series.

Example 3. Applying the Absolute Convergence Test

1. $\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n\ln n}{n^2}$ converges absolutely
2. $\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n\ln n}{n}$ converges conditionally

Example 4. Being Alternating is Good for Convergence

1. If $p$ is a positive constant, then

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}\text{ converges when }p>1,\text{but }\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^p}\text{ converges when }p>0.$$

$\sum _n\frac{(-1{)}^n}{n^p}$	$0<p\le 1$	$p>1$
Converge	✔️	✔️
Absolutely	❌	✔️
Conditionally	✔️	❌

The difference between absolute and conditional convergences is best shown from the following phenomenon.

$$\text{Absolute Convergent Series: Rearrange the terms does not change the sum}\text{Conditional Convergent Series: Rearrange the terms DOES change the sum!}$$

Rearranging Series

Theorem 17. The Rearrangement Theorem for Absolutely Convergent Series

If $\sum _n{a}_n$ converges absolutely, and $b_1,b_2,b_3,\cdot ,b_n,\cdots$ is any rearrangement of the sequence $a_n$, then $\sum _n{b}_n$ converges absolutely and

$$\sum _n{b}_n=\sum _n{a}_n.$$

Example 5. Rearrangement on Absolutely Convergent Series

$$1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\cdots$$

Can be rearranged to

$$1-\frac{1}{4}-\frac{1}{16}+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}-\frac{1}{36}-\frac{1}{64}-\frac{1}{100}-\frac{1}{144}+\cdots$$

And since the series absolutely convergent, it really doesn’t matter how we rearrange it. It keeps absolutely converging to the same value.

However, a rearrangement on a conditionally convergent series is more interesting!

Example 6. Rearranging the Alternating Harmonic Series

The alternating harmonic series is conditionally convergent series! Thus

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots$$

can be rearranged to reach different sums.

• Rearrange to $1/2$ of the original sum. Consider the following rearrangement

  $$(1-\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{6}-\frac{1}{8})+(\frac{1}{5}-\frac{1}{10}-\frac{1}{12})+\cdots +(\frac{1}{2n+1}-\frac{1}{4n+2}-\frac{1}{4n+4})+\cdots$$

  Note that the grouping partial sum has the following form

  $$\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{2n+1}-\frac{1}{4n+2}-\frac{1}{4n+4})=\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{4n+2}-\frac{1}{4n+4})=\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{2n+1}-\frac{1}{2n+2})$$

• Rearrange to a divergent series. Since series of terms $\sum \frac{1}{2n-1}$ diverges to $+\mathrm{\infty }$, and the series of terms $\sum \frac{-1}{2n}$ diverges to $-\mathrm{\infty }$, we construct the following rearrangement

  $$ \scriptsize\underbrace{\overbrace{\underbrace{1+\frac13 + \frac15 + \cdots + \frac1{2m_1+1}}{\text{Larger than } 2} - \frac1{2}-\frac14 - \cdots - \frac{1}{2n_1 +1}}^{\text{smaller than }-3}+\frac{1}{2m_1+3}+\frac1{2m_1+5} + \cdots + \frac1{2m_2+1}}{\text{larger than }4}+\cdots $$

  such that the partial sum swings large in either direction

• Rearrange to converge to 1. Making the partial sum swings about 1

  • Adds 1 (Making the partial sum just $\ge 1$)
  • Subtracts $\frac{1}{2}$ (Making the partial sum just down to a value < 1)
  • Adds $\frac{1}{3}$, $\frac{1}{5}$ (Making the partial sum just >1 again)
  • Subtracts $\frac{1}{4}$ (Making the partial sum just < 1 again)
  • Adds $\frac{1}{7}$, $\frac{1}{9}$ (Making the partial sum just > 1 again)
  • …

  The new arrangement looks like

  $$1-\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{4}+\frac{1}{7}+\frac{1}{9}-\frac{1}{6}+\frac{1}{11}+\frac{1}{13}-\frac{1}{8}+\frac{1}{15}+\frac{1}{17}-\frac{1}{10}+\frac{1}{19}+\frac{1}{21}-\frac{1}{12}+\frac{1}{23}+\frac{1}{25}-\frac{1}{14}+\frac{1}{27}-\frac{1}{16}+\cdots$$

  Since the amount by which our partial sums exceed 1 or fall below it approaches zero, the new series converges to 1.

Summary