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Next: 11.4 Absolute Convergence Up: 11. Infinite Series Previous: 11.2 Convergence Tests   Index

11.3 Alternating Series

11.29   Definition (Alternating series.) Series of the form $\displaystyle {\sum\{(-1)^na_n\}}$ or $\displaystyle {\sum\{(-1)^{n+1}a_n\}}$ where $a_j\geq 0$ for all $j$ are called alternating series.

11.30   Theorem (Alternating series test.) Let $f$ be a decreasing sequence of positive numbers such that $\{f(n)\}\to 0$. Then $\{(-1)^nf(n)\}$ is summable. Moreover,



\begin{displaymath}\vert\sum_{j=0}^n (-1)^jf(j)-\sum_{j=0}^\infty(-1)^jf(j)\vert\leq f(n+1)\end{displaymath}

for all $m,n\in\mbox{{\bf N}}$.

Proof: Let $\displaystyle {S_n=\sum_{j=0}^n(-1)^j f(j)}$. For all $n\in\mbox{{\bf N}}$,

\begin{displaymath}S_{2(n+1)}=S_{2n+2}=S_{2n}-f(2n+1) + f(2n+2)\leq S_{2n}\end{displaymath}


\begin{displaymath}S_{2(n+1)+1}=S_{2n+1}+f(2n+2)-f(2n+3)\geq S_{2n+1}.\end{displaymath}

Thus $\{S_{2n}\}$ is decreasing and $\{S_{2n+1}\}$ is increasing. Also, for all $n\in\mbox{{\bf N}}$,

\begin{displaymath}S_1\leq S_{2n+1}=S_{2n}-f(2n+1)\leq S_{2n}\end{displaymath}

so $\{S_{2n}\}$ is bounded below by $S_1$, and

\begin{displaymath}S_{2n+1}=S_{2n}-f(2n+1)\leq S_{2n}\leq S_0\end{displaymath}

so $\{S_{2n+1}\}$ is bounded above by $S_0$.

It follows that there exist real numbers $L$ and $M$ such that

&\;&\{S_{2n}\}\to L\mbox{ and }S_{2n}\geq L\mbox{ for all }n\i...
...o M\mbox{ and }S_{2n+1}\leq M\mbox{ for all }n\in\mbox{{\bf N}}.



so $L=M$.

It follows from the next lemma that $\{S_n\}\to L$; i.e.,

\begin{displaymath}\displaystyle {M=L=\lim

Since for all $n\in\mbox{{\bf N}}$

\begin{displaymath}S_{2n+1}\leq L\leq S_{2n},\end{displaymath}

we have

\begin{displaymath}\vert L-S_{2n}\vert\leq S_{2n}-S_{2n+1}=f(2n+1)\end{displaymath}

and since

S_{2n+1}\leq L&\leq& S_{2n+2}\\
\vert L-S_{2n+1}\vert&\leq&S_{2n+2}-S_{2n+1}=f(2n+2).

Thus, in all cases, $\vert L-S_n\vert<f(n+1)$; i.e., $\displaystyle {\sum_{j=0}^n(-1)^jf(j)}$ approximates $\displaystyle {\sum_{j=0}^\infty(-1)^jf(j)}$ with an error of no more than $f(n+1)$. $\mid\!\mid\!\mid$

11.31   Lemma. Let $\{a_n\}$ be a real sequence and let $L\in\mbox{{\bf R}}$. Suppose $\{a_{2n}\}\to L$ and $\{a_{2n+1}\}\to L$. Then $\{a_n\}\to L$.

Proof: Let $N$ be a precision function for $\{a_{2n}-L\}$ and let $M$ be a precision function for $\{a_{2n+1}-L\}$. For all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$, define

\begin{displaymath}N_{a-\tilde L}(\varepsilon)=\max\left(2N(\varepsilon),2M(\varepsilon)+1\right).\end{displaymath}

I claim $N_{a-\tilde L}$ is a precision function for $a-\tilde L$, and hence $a\to
L$. Let $n\in\mbox{{\bf N}}$.
Case 1:
$n$ is even. Suppose $n$ is even. Say $n=2k$ where $k \in\mbox{{\bf N}}$. Then

(n\geq N_{a-\tilde L}(\varepsilon) &\mbox{$\Longrightarrow$}&
&\mbox{$\Longrightarrow$}&\vert a_n-L\vert<\varepsilon,

Case 2:
$n$ is odd. Suppose $n$ is odd. Say $n=2k+1$ where $k \in\mbox{{\bf N}}$. Then

(n\geq N_{a-\tilde L}(\varepsilon) &\mbox{$\Longrightarrow$}&
&\mbox{$\Longrightarrow$}& \vert a_n-L\vert<\varepsilon.

Hence, in all cases,

\begin{displaymath}n\geq N_{a-\tilde L}(\varepsilon)\mbox{$\hspace{1ex}\Longrigh...
...e{1ex}$}\vert a_n-L\vert<\varepsilon.\mbox{ $\mid\!\mid\!\mid$}\end{displaymath}

11.32   Remark. The alternating series test has obvious generalizations for series such as

\begin{displaymath}\sum\{(-1)^{j+1}f(j)\}\mbox{ or }\sum\{(-1)^jf(j)\}_{j\geq k},\end{displaymath}

and we will use these generalizations.

11.33   Example. If $0\leq t\leq 1$, then

\begin{displaymath}\left\{ {{t^{2n}}\over {(2n)!}}\right\} \mbox{ and } \left\{ {{t^{2n+1}}\over

are decreasing positive null sequences, so

\begin{displaymath}\left\{ {{(-1)^nt^{2n}}\over {(2n)!}}\right\}\mbox{ and }\left\{ {{(-1)^nt^{2n+1}}\over

are summable; i.e.,

{(2j+1)!}}\right\}\mbox{ converge. }\end{displaymath}

(These are the sequences we called $\{C_n(t)\}$ and $\{S_n(t)\}$ in example 10.3.)

Also, $\displaystyle {\sum_{j=0}^\infty{{(-1)^j\left({1\over {10}}\right)^{2j}}\over
{(2j)!}}=1-{1\over {200}}+{1\over {240000}}}$, with an error smaller than
$\displaystyle {
{1\over {720000000}}}$. My calculator says



1-{1\over {200}}+{1\over {240000}}=0.995004166.

11.34   Entertainment. Since $\displaystyle {\left\{ {{t^n}\over n}\right\}_{n\geq 1}}$ is a decreasing positive null sequence for $0\leq t\leq 1$, it follows that $\displaystyle {\sum\left\{{{(-1)^{n-1}t^n}\over
n}\right\}_{n\geq 1}}$ converges for $0\leq t\leq 1$. We will now explicitly calculate the limit of this series using a few ideas that are not justified by results proved in this course. We know that for all $x\in\mbox{{\bf R}}\setminus \{-1\}$, and all $n\in\mbox{{\bf N}}$,

\begin{displaymath}1-x+x^2-x^3+\cdots +(-x)^{n-1}={{1-(-x)^n}\over {1-(-x)}}={1\over
{1+x}}+(-1)^{n+1}{{x^n}\over {1+x}}.\end{displaymath}

Hence, for all $t>-1$,

\begin{displaymath}\int_0^t 1-x+x^2+\cdots +(-x)^{n-1}dx=\int_0^t{1\over
{1+x}}dx+(-1)^{n+1}\int_0^t{{x^n}\over {1+x}}dx;\end{displaymath}


\begin{displaymath}x-{{x^2}\over 2}+{{x^3}\over 3}+\cdots+{{(-1)^{n-1}x^n}\over
...=\ln(1+x)\Big\vert _0^t+(-1)^{n+1}\int_0^t{{x^n}\over {1+x}}dx.\end{displaymath}


\begin{displaymath}t-{{t^2}\over 2}+{{t^3}\over 3}+\cdots+{{(-1)^{n-1}t^n}\over
n}=\ln(1+t)+(-1)^{n+1}\int_0^t{{x^n}\over {1+x}}dx.\end{displaymath}


\begin{displaymath}\left\vert t-{{t^2}\over 2}+{{t^3}\over 3}+\cdots+{{(-1)^{n-1...
...)\right\vert=\left\vert \int_0^t{{x^n}\over {1+x}}dx\right\vert\end{displaymath}

for all $t>-1$.

If we can show that $\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$ is a null sequence, it follows that

\begin{displaymath}\left\{t-{{t^2}\over 2}+{{t^3}\over 3}+\cdots+{{(-1)^{n-1}t^n}\over

or in other words,
\ln(1+t)=\sum_{j=1}^\infty{{(-1)^{j+1}t^j}\over j}.
\end{displaymath} (11.35)

I claim $\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$ is a null sequence for $-1<t\leq 1$ and hence (11.35) holds for $-1<t\leq 1$. In particular,

\begin{displaymath}\ln(2)=1-{1\over 2}+{1\over 3}-{1\over 4}+{1\over 5}-{1\over 6}+\cdots.\end{displaymath}

First suppose $t\geq 0$, then $\displaystyle {{1\over {1+x}}x^n\leq x^n}$ for $0\leq x\leq t$, so

\begin{displaymath}0=\int_0^t{1\over {1+x}}\cdot x^n dx\leq\int_0^tx^n dx=\left. {{x^{n+1}}\over
{n+1}}\right\vert _0^t={{t^{n+1}}\over {n+1}}.\end{displaymath}

Since $\displaystyle { \left\{ {{t^{n+1}}\over {n+1}}\right\}}$ is a null sequence for $0\leq t\leq 1$, it follows from the comparison test that $\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$ is a null sequence for $0\leq t\leq 1$. Now suppose $-1<t<0$. Then

\begin{displaymath}{1\over {1+x}}\leq {1\over {1+t}}\mbox{ for } t\leq x\leq 0,\end{displaymath}

so $\displaystyle { {{\vert x\vert^n}\over {1+x}}\leq {{\vert x\vert^n}\over {1+t}}}$ and

\left\vert \int_0^t{{x^n}\over {1+x}}dx\right\vert &=& \int_t^...
...n dx\\
&=&{1\over {1+t}}\cdot{{\vert t\vert^{n+1}}\over {n+1}}.

If $-1<t<0$, then $\displaystyle { \left\{{1\over {1+t}} \cdot{{\vert t\vert^{n+1}}\over {n+1}}\right\}}$ is a null sequence, so $\displaystyle {\left\{\int_0^t{{x^n}\over {1+x}}dx\right\}}$ is a null sequence. $\mid\!\mid\!\mid$

11.36   Entertainment. By starting with the formula

\begin{displaymath}1-x^2+x^4-x^6+\cdots+(-x^2)^{n-1}={{1-(-x^2)^n}\over {1-(-x^2)}}\end{displaymath}

for all $x\in\mbox{{\bf R}}$ and using the ideas from the last example, show that
\sum_{j=0}^\infty {{(-1)^jx^{2j+1}}\over {(2j+1)}}=\arctan(x)\mbox{ for all }x\in[-1,1].
\end{displaymath} (11.37)

Conclude that

\begin{displaymath}\displaystyle { {\pi\over 4}=1-{1\over 3}+{1\over 5}-{1\over 7}+{1\over 9}-{1\over

11.38   Exercise. Determine whether or not the following series converge.
$\displaystyle {\sum\left\{ (-1)^n{{n+1}\over {n^2}}\right\}_{n\geq 1}}$
$\displaystyle {\sum\left\{ (-1)^n{{n+1}\over n}\right\}_{n\geq 1}}$
$\displaystyle {\sum\left\{ {{(-1)^nt^{2n}}\over {(2n)!}}\right\}}$ (assume here $-1\leq t\leq 1$).

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Next: 11.4 Absolute Convergence Up: 11. Infinite Series Previous: 11.2 Convergence Tests   Index