11.3 Alternating Series

11.29   Definition (Alternating series.) Series of the form $${\sum\{(-1)^na_n\}}$$ or $${\sum\{(-1)^{n+1}a_n\}}$$ 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,

$$\sum_{j=0}^{2m+1}(-1)^j f(j)\leq\sum_{j=0}^\infty(-1)^jf(j)\leq\sum_{j=0}^{2n}(-1)^jf(j)$$

and

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

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

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

$$S_{2(n+1)}=S_{2n+2}=S_{2n}-f(2n+1) + f(2n+2)\leq S_{2n}$$

and

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

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

$$S_1\leq S_{2n+1}=S_{2n}-f(2n+1)\leq S_{2n}$$

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

$$S_{2n+1}=S_{2n}-f(2n+1)\leq S_{2n}\leq S_0$$

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

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

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

Now

$$\begin{eqnarray*} L-M&=&\lim\{S_{2n}\}-\lim\{S_{2n+1}\}=\lim\{S_{2n}-S_{2n+1}\}\\ &=&\lim\{f(2n+1)\}=0, \end{eqnarray*}$$

so $L=M$.

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

$$\displaystyle {M=L=\lim S_n=\sum_{n=0}^\infty(-1)^nf(n)}.$$

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

$$S_{2n+1}\leq L\leq S_{2n},$$

we have

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

and since

$$\begin{eqnarray*} S_{2n+1}\leq L&\leq& S_{2n+2}\\ \vert L-S_{2n+1}\vert&\leq&S_{2n+2}-S_{2n+1}=f(2n+2). \end{eqnarray*}$$

Thus, in all cases, $\vert L-S_n\vert<f(n+1)$; i.e., $${\sum_{j=0}^n(-1)^jf(j)}$$ approximates $${\sum_{j=0}^\infty(-1)^jf(j)}$$ 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

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

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

$$\begin{eqnarray*} (n\geq N_{a-\tilde L}(\varepsilon) &\mbox{$\Longrightarrow$}& ... ...silon\\ &\mbox{$\Longrightarrow$}&\vert a_n-L\vert<\varepsilon, \end{eqnarray*}$$

Case 2:

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

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

Hence, in all cases,

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

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

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

and we will use these generalizations.

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

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

are decreasing positive null sequences, so

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

are summable; i.e.,

$$\left\{\sum_{j=0}^n{{(-1)^jt^{2j}}\over {(2j)!}}\right\}\mbox... ...0}^n{{(-1)^jt^{2j+1}}\over {(2j+1)!}}\right\}\mbox{ converge. }$$

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

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

$$\cos(.1)=0.995004165$$

and

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

11.34   Entertainment. Since $${\left\{ {{t^n}\over n}\right\}_{n\geq 1}}$$ is a decreasing positive null sequence for $0\leq t\leq 1$, it follows that $${\sum\left\{{{(-1)^{n-1}t^n}\over n}\right\}_{n\geq 1}}$$ 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 $n\in\mbox{{\bf N}}$,

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

Hence, for all $t>-1$,

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

i.e.,

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

Thus

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

Hence

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

for all $t>-1$.

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

$$\left\{t-{{t^2}\over 2}+{{t^3}\over 3}+\cdots+{{(-1)^{n-1}t^n}\over n}\right\}\to\ln(1+t),$$

or in other words,

$$\ln(1+t)=\sum_{j=1}^\infty{{(-1)^{j+1}t^j}\over j}.$$ (11.35)

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

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

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

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

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

$${1\over {1+x}}\leq {1\over {1+t}}\mbox{ for } t\leq x\leq 0,$$

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

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

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

11.36   Entertainment. By starting with the formula

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

for all $x\in\mbox{{\bf R}}$ and using the ideas from the last example, show that

$$\sum_{j=0}^\infty {{(-1)^jx^{2j+1}}\over {(2j+1)}}=\arctan(x)\mbox{ for all }x\in[-1,1].$$ (11.37)

Conclude that

$$\displaystyle { {\pi\over 4}=1-{1\over 3}+{1\over 5}-{1\over 7}+{1\over 9}-{1\over {11}}+\cdots}$$

11.38   Exercise. Determine whether or not the following series converge.

a)

$${\sum\left\{ (-1)^n{{n+1}\over {n^2}}\right\}_{n\geq 1}}$$

b)

$${\sum\left\{ (-1)^n{{n+1}\over n}\right\}_{n\geq 1}}$$

c)

$${\sum\left\{ {{(-1)^nt^{2n}}\over {(2n)!}}\right\}}$$ (assume here $-1\leq t\leq 1$).