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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
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\lhead{\color{blue} \Large Math 208}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\LARGE Series 4 - Alternating Series}
\rfoot{\small Last revised:April 25, 2013}
\rhead{\large  page \thepage }
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\begin{document}


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Theorem: \ (The Alternating Series Test or Leibniz's Test) \ The series%
\begin{equation*}
\dsum\limits_{n=1}^{\infty }\left( -1\right)
^{n+1}u_{n}=u_{1}-u_{2}+u_{3}-u_{4}.....
\end{equation*}%
converges if all three of the following conditions are satisfied:\newline
1) \ $u_{n}$ is positive for all $n$.\newline
2) \ $u_{n}$ are (eventually) non-increasing: $u_{n}\geq u_{n+1}$ for all $%
n\geq N$ for some integer $N$.\newline
3) \ $\lim\limits_{n\rightarrow \infty }u_{n}=0$ \ 
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\medskip

\bigskip

Proof: \ Assume that $N=1$. \ If $n$ is even, say $n=2k$ then the partial
sum $s_{n}$ is%
\begin{equation*}
s_{n}=\left( u_{1}-u_{2}\right) +\left( u_{3}-u_{4}\right) +...+\left(
u_{2k-1}-u_{2k}\right)
\end{equation*}%
Since $\left\{ u_{n}\right\} $ is non-increasing, each difference $%
u_{j}-u_{j+1}$ is non-negative, thus $s_{n}$ is non-negative and
non-decreasing. \ Let us look at $s_{n}$ again, but this time as%
\begin{equation*}
s_{n}=u_{1}-\left( u_{2}-u_{3}\right) -\left( u_{4}-u_{5}\right) -...-\left(
u_{2k-2}-u_{2k-1}\right) -u_{2k}
\end{equation*}

This line shows that $s_{n}\leq u_{1}$. \ \ Thus the sequence $\left\{
s_{n}\right\} =\left\{ s_{2k}\right\} $ of even partial sums is bounded from
above (one upper bound is $u_{1}$). \ Because $\left\{ s_{n}\right\}
=\left\{ s_{2k}\right\} $ is a sequence that is non-decreasing and bounded
from above, it is convergent. \ Let us denote the limit by $L$. \ \bigskip

Let us now consider the odd partial sums $\left\{ s_{2k+1}\right\}
=s_{2k}+u_{2k+1}$. \ Recall that $\lim\limits_{n\rightarrow \infty }u_{n}=0$%
. \ Then%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }s_{2k+1}=\lim\limits_{n\rightarrow \infty
}s_{2k}+u_{2k+1}=\lim\limits_{n\rightarrow \infty
}s_{2k}+\lim\limits_{n\rightarrow \infty }u_{2k+1}=L+0=L
\end{equation*}%
This completes our proof.\bigskip

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Theorem: \ (The Alternating Series Estimation Theorem) \ If an alternating
series $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}u_{n}$\ satisfies
Leibniz's test, then 
\begin{equation*}
s_{n}=u_{1}-u_{2}+u_{3}-u_{4}+....+\left( -1\right) ^{n+1}u_{n}
\end{equation*}%
approximates the sum $S$ of the series with an error whose absolute value is
less than $u_{n+1}$, the absolute value of the first unused term. \
Furthermore, the sum $S$ lies between any two successive partial sums $s_{n}$
and $s_{n+1}$, and the remainder, $S-s_{n}$ has the same sign as the first
unused term.%
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\bigskip

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Definition: \ A series $\dsum a_{n}$\ \textbf{converges absolutely} if the
corresponding series of absolute values $\dsum \left\vert a_{n}\right\vert $
converges.%
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\bigskip

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Definition: \ A series that\ converges but does not converge absolutely 
\textbf{converges conditionally}.%
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\bigskip

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Theorem: \ (The Absolute Convergence Test) \ If $\dsum\limits_{n=1}^{\infty
}\left\vert a_{n}\right\vert $\ converges, then $\dsum\limits_{n=1}^{\infty
}a_{n}$ converges.%
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\bigskip

Proof: \ Suppose that $\dsum\limits_{n=1}^{\infty }\left\vert
a_{n}\right\vert $ converges. \ Then the series $\dsum\limits_{n=1}^{\infty
}2\left\vert a_{n}\right\vert $ also converges. \ \ Consider now the series $%
\dsum\limits_{n=1}^{\infty }\left( \left\vert a_{n}\right\vert +a_{n}\right) 
$. \ We claim that this series converges as well, by the comparison test:
for all $n\in 
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$,%
\begin{equation*}
0\leq \left\vert a_{n}\right\vert +a_{n}\leq 2\left\vert a_{n}\right\vert
\end{equation*}%
So $\dsum\limits_{n=1}^{\infty }\left( \left\vert a_{n}\right\vert
+a_{n}\right) $ also converges. \ Now consider the difference $%
\dsum\limits_{n=1}^{\infty }\left( \left\vert a_{n}\right\vert +a_{n}\right)
-\left\vert a_{n}\right\vert $. \ Since this is the difference of two
convergent series, it also converges. \ This completes our proof. \ \bigskip

Theorem: \ (The Rearrangement Theorem for Absolutely Convergent Series) \ If 
$\dsum\limits_{n=1}^{\infty }a_{n}$\ converges absolutely, and $b_{1}$,$%
b_{2} $,$b_{3}$,$...$,$b_{n},.....$ is any rearrangment of the sequence $%
\left\{ a_{n}\right\} $ then $\dsum\limits_{n=1}^{\infty }b_{n}$ converges
absolutely and%
\begin{equation*}
\dsum\limits_{n=1}^{\infty }b_{n}=\dsum\limits_{n=1}^{\infty }a_{n}
\end{equation*}

Note that this is not true for conditionally convergent series.\bigskip
\bigskip \bigskip

\begin{center}
{\LARGE Sample Problems\bigskip }
\end{center}

\begin{enumerate}
\item Determine convergence or divergence of the alternative series.%
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a) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{1}{\sqrt{n}}%
\medskip \medskip $

b) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{n^{2}+5}{%
n^{2}+4}\medskip \medskip $

c) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\dfrac{n}{n^{2}-1}%
\medskip \medskip $

d) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\dfrac{10^{n}}{\left(
n+1\right) !}\medskip \medskip $

e) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\ln \left( 1+\dfrac{1%
}{n}\right) \medskip \medskip $

f) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{\tan ^{-1}n}{%
n^{2}+1}\medskip \medskip $

g) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{n}{n+1}%
\medskip \medskip $

h) $\ \dsum\limits_{n=1}^{\infty }\left( -2\right) ^{-n}$ $\medskip \medskip 
$\ 
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\item Determine whether the following series converge absolutely, converge
conditionally, or diverge.%
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a) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\left( -100\right) ^{n}}{n!}%
\medskip \medskip $

b) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{1}{\sqrt{n}}%
\medskip \medskip $

c) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\cos n\pi }{n\sqrt{n}}\medskip
\medskip $

d) $\ \dsum\limits_{n=2}^{\infty }\left( -1\right) ^{n}\dfrac{1}{n\ln n}%
\medskip \medskip $

e) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n+1}\left(
n!\right) ^{2}}{\left( 2n\right) !}\medskip \medskip $

f) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n}\left(
n!\right) ^{2}3^{n}}{\left( 2n+1\right) !}\medskip \medskip $

f) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{n}{n+1}%
\medskip \medskip $

g) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\left( \sqrt{n+1}-%
\sqrt{n}\right) \medskip \medskip $ \ 
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\end{enumerate}

\pagebreak

\begin{center}
{\LARGE Solutions - Sample Problems\bigskip }
\end{center}

\begin{enumerate}
\item Determine convergence or divergence of the alternative series.%
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a) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{1}{\sqrt{n}}$

converges since $\left\{ a_{n}\right\} $ is alternating, decreasing, and \
approaches zero

b) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{n^{2}+5}{%
n^{2}+4}$

diverges since $a_{n}$ fails to approach zero

c) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\dfrac{n}{n^{2}-1}$

converges since $\left\{ a_{n}\right\} $ is alternating, decreasing, and \
approaches zero

d) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\dfrac{10^{n}}{\left(
n+1\right) !}$

converges since $\left\{ a_{n}\right\} $ is alternating, decreasing after $%
N=11$, and \ approaches zero

e) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\ln \left( 1+\dfrac{1%
}{n}\right) $

converges since $\left\{ a_{n}\right\} $ is alternating, decreasing, and \
approaches zero

f) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{\tan ^{-1}n}{%
n^{2}+1}$

converges since $\left\{ a_{n}\right\} $ is alternating, decreasing, and \
approaches zero

g) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{n}{n+1}$ \ \
\ \ diverges since $a_{n}$ fails to approach zero

h) $\ \dsum\limits_{n=1}^{\infty }\left( -2\right) ^{-n}$ \ \ \ \ converges
since it is a geometric series with $r=-\dfrac{1}{2}$

\item Determine whether the following series converge absolutely, converge
conditionally, or diverge.%
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a) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\left( -100\right) ^{n}}{n!}$

The series converges absolutely. \ We prove it using the ratio test%
\begin{eqnarray*}
\rho &=&\lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}%
\right\vert =\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{100^{n+1}}{%
\left( n+1\right) !}}{\dfrac{100^{n}}{n!}}=\lim\limits_{n\rightarrow \infty
}\left( \dfrac{100^{n+1}}{\left( n+1\right) !}\cdot \dfrac{n!}{100^{n}}%
\right) =\lim\limits_{n\rightarrow \infty }\left( \dfrac{100\cdot 100^{n}}{%
\left( n+1\right) n!}\cdot \dfrac{n!}{100^{n}}\right) \\
&=&\lim\limits_{n\rightarrow \infty }\left( \dfrac{100}{n+1}\right) =0<1
\end{eqnarray*}%
\pagebreak

b) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{1}{\sqrt{n}}$

This series converges conditionally. \ $\dfrac{1}{\sqrt{n}}$ is decrasing
and approaches zero. \ However, it is not absolutely convergent, because $%
\dsum\limits_{n=1}^{\infty }\dfrac{1}{\sqrt{n}}$ diverges by the integral
test.

c) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\cos n\pi }{n\sqrt{n}}$

This series converges absolutely because $\dsum\limits_{n=1}^{\infty
}\left\vert \dfrac{\cos n\pi }{n\sqrt{n}}\right\vert
=\dsum\limits_{n=1}^{\infty }\dfrac{1}{n\sqrt{n}}$ converges by the integral
test

d) $\ \dsum\limits_{n=2}^{\infty }\left( -1\right) ^{n}\dfrac{1}{n\ln n}$

This series converges conditionally because it satisfies Leibniz's test. \
However, the series $\dsum\limits_{n=2}^{\infty }\left\vert a_{n}\right\vert
=\dsum\limits_{n=2}^{\infty }\dfrac{1}{n\ln n}$ diverges by the integral test

e) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n+1}\left(
n!\right) ^{2}}{\left( 2n\right) !}$

This series converges absolutely because $\dsum\limits_{n=1}^{\infty
}\left\vert \dfrac{\left( -1\right) ^{n+1}\left( n!\right) ^{2}}{\left(
2n\right) !}\right\vert =\dsum\limits_{n=1}^{\infty }\dfrac{\left( n!\right)
^{2}}{\left( 2n\right) !}$ converges by the ratio test%
\begin{eqnarray*}
\rho &=&\lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}%
\right\vert =\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{\left( \left(
n+1\right) !\right) ^{2}}{\left( 2\left( n+1\right) \right) !}}{\dfrac{%
\left( n!\right) ^{2}}{\left( 2n\right) !}}=\lim\limits_{n\rightarrow \infty
}\dfrac{\dfrac{\left( \left( n+1\right) n!\right) ^{2}}{\left( 2n+2\right) !}%
}{\dfrac{\left( n!\right) ^{2}}{\left( 2n\right) !}}=\lim\limits_{n%
\rightarrow \infty }\dfrac{\left( n+1\right) ^{2}\left( n!\right) ^{2}}{%
\left( 2n+2\right) \left( 2n+1\right) \left( 2n\right) !}\cdot \dfrac{\left(
2n\right) !}{\left( n!\right) ^{2}} \\
&=&\lim\limits_{n\rightarrow \infty }\dfrac{\left( n+1\right) ^{2}}{2\left(
n+1\right) \left( 2n+1\right) }=\lim\limits_{n\rightarrow \infty }\dfrac{n+1%
}{2\left( 2n+1\right) }=\lim\limits_{n\rightarrow \infty }\dfrac{1+\dfrac{1}{%
n}}{2\left( 2+\dfrac{1}{n}\right) }=\dfrac{1}{4}<1
\end{eqnarray*}

f) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n}\left(
n!\right) ^{2}3^{n}}{\left( 2n+1\right) !}$ \ This series converges
absolutely because of the ratio test%
\begin{eqnarray*}
\rho &=&\lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}%
\right\vert =\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{\left( \left(
n+1\right) !\right) ^{2}3^{n+1}}{\left( 2\left( n+1\right) +1\right) !}}{%
\dfrac{\left( n!\right) ^{2}3^{n}}{\left( 2n+1\right) !}}=\lim\limits_{n%
\rightarrow \infty }\dfrac{\left( \left( n+1\right) n!\right) ^{2}3^{n}\cdot
3}{\left( 2n+3\right) !}\cdot \dfrac{\left( 2n+1\right) !}{\left( n!\right)
^{2}3^{n}} \\
&=&\lim\limits_{n\rightarrow \infty }\dfrac{3\left( n+1\right) ^{2}\left(
n!\right) ^{2}\left( 2n+1\right) !}{\left( 2n+3\right) \left( 2n+2\right)
\left( 2n+1\right) !\left( n!\right) ^{2}}=\lim\limits_{n\rightarrow \infty }%
\dfrac{3\left( n+1\right) ^{2}}{\left( 2n+3\right) \left( 2n+2\right) }%
=\lim\limits_{n\rightarrow \infty }\dfrac{3\left( n+1\right) ^{2}}{\left(
2n+3\right) 2\left( n+1\right) } \\
&=&\lim\limits_{n\rightarrow \infty }\dfrac{3\left( n+1\right) }{2\left(
2n+3\right) }=\dfrac{3}{4}<1
\end{eqnarray*}

f) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n+1}\dfrac{n}{n+1}$ \ \
\ \ This series diverges because $a_{n}$ fails to approach zero.

g) \ $\dsum\limits_{n=1}^{\infty }\left( -1\right) ^{n}\left( \sqrt{n+1}-%
\sqrt{n}\right) $

This series converges conditionally, because it satisfies Leibniz's test:%
\begin{eqnarray*}
\lim\limits_{n\rightarrow \infty }\left\vert a_{n}\right\vert
&=&\lim\limits_{n\rightarrow \infty }\left( \sqrt{n+1}-\sqrt{n}\right)
=\lim\limits_{n\rightarrow \infty }\left( \left( \sqrt{n+1}-\sqrt{n}\right) 
\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\right)
=\lim\limits_{n\rightarrow \infty }\left( \dfrac{n+1-n}{\sqrt{n+1}+\sqrt{n}}%
\right) \\
&=&\lim\limits_{n\rightarrow \infty }\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=0
\end{eqnarray*}

However, the series $\dsum\limits_{n=1}^{\infty }\left\vert a_{n}\right\vert
=\dsum\limits_{n=1}^{\infty }\left( \sqrt{n+1}-\sqrt{n}\right) $ diverges by
the comparison test:%
\begin{equation*}
\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\geq \dfrac{1}{\sqrt{n+1}+%
\sqrt{n+1}}=\dfrac{1}{2\sqrt{n}}
\end{equation*}%
and $\dsum\limits_{n=1}^{\infty }\dfrac{1}{2\sqrt{n}}$ is divergent.
\end{enumerate}

\bigskip

\vspace{3in}

\vspace{2in}

\vspace{1in}

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{\small 
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\href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{%
For more documents like this, visit our page at\
https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail
questions or comments to mhidegkuti@ccc.edu.}

\end{document}