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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}
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\newtheorem{exercise}[theorem]{Exercise}
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\lhead{\color{blue} \large Math 208}
\lfoot{\small   \copyright $\;$  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\Large Series 1}
\rfoot{\small Last revised: April 9, 2013}
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\begin{document}


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Definition: \ Given a sequence of numbers $\left\{ a_{n}\right\} $, the
expression 
\begin{equation*}
a_{1}+a_{2}+a_{3}+...+a_{n}+....
\end{equation*}%
is called an \textbf{infinite series}. \ The number $a_{n}$ is the $%
\boldsymbol{n}$\textbf{th term} of the series. \ The sequence $\left\{
s_{n}\right\} $ defined by 
\begin{eqnarray*}
s_{1} &=&a_{1} \\
s_{2} &=&a_{1}+a_{2} \\
s_{3} &=&a_{1}+a_{2}+a_{3} \\
&&\vdots \\
s_{n} &=&a_{1}+a_{2}+a_{3}+...+a_{n}=\dsum\limits_{k=1}^{n}a_{k}
\end{eqnarray*}%
is called the \textbf{sequence of partial sums} of the series, the number $%
s_{n}$ being the $\boldsymbol{n}$\textbf{th partial sum}. \ If the sequence
of partial sums converges to a limit $L,$ we say that the series converges
and that its \textbf{sum} is $L$. In this case, we also write 
\begin{equation*}
a_{1}+a_{2}+a_{3}+...+a_{n}+....=\dsum\limits_{n=1}^{\infty }a_{n}=L
\end{equation*}%
If the sequence of partial sums of the series does not converge, we say that
the series \textbf{diverges}.%
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\medskip

\bigskip

Example 1) \ $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n\left( n+1\right) }$ is
convergent.\bigskip

Proof: \ The series is $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{%
3\cdot 4}+....$ \ Consider the sequence $\left\{ s_{n}\right\} $ of partial
sums.%
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\begin{eqnarray*}
s_{1} &=&\dfrac{1}{2} \\
s_{2} &=&\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3} \\
s_{3} &=&\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{4}
\end{eqnarray*}

\begin{eqnarray*}
s_{4} &=&\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}=\dfrac{4}{5}
\\
s_{5} &=&\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=%
\dfrac{5}{6} \\
s_{6} &=&\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+%
\dfrac{1}{42}=\dfrac{6}{7}
\end{eqnarray*}%
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\bigskip

It appears that $s_{n}=\dfrac{n}{n+1}$. \ If that was so, then the sum of
the series can be easily found as the limit of the sequence of partial sums:%
\begin{equation*}
s=\lim\limits_{n\rightarrow \infty }\dfrac{n}{n+1}=\lim\limits_{n\rightarrow
\infty }\left( 1-\dfrac{1}{n+1}\right) =1
\end{equation*}%
This is the case indeed. \ We can prove it in general using partial
fractions:%
\begin{equation*}
a_{n}=\dfrac{1}{n\left( n+1\right) }=\dfrac{1}{n}-\dfrac{1}{n+1}
\end{equation*}%
and so%
\begin{eqnarray*}
s_{1} &=&a_{1}=\dfrac{1}{1}-\dfrac{1}{2}=1-\dfrac{1}{2} \\
s_{2} &=&a_{1}+a_{2}=\left( \dfrac{1}{1}-\dfrac{1}{2}\right) +\left( \dfrac{1%
}{2}-\dfrac{1}{3}\right) =1-\dfrac{1}{3} \\
s_{3} &=&a_{1}+a_{2}+a_{3}=\left( \dfrac{1}{1}-\dfrac{1}{2}\right) +\left( 
\dfrac{1}{2}-\dfrac{1}{3}\right) +\left( \dfrac{1}{3}-\dfrac{1}{4}\right) =1-%
\dfrac{1}{4} \\
s_{n} &=&a_{1}+a_{2}+...+a_{n}=\left( \dfrac{1}{1}-\dfrac{1}{2}\right)
+\left( \dfrac{1}{2}-\dfrac{1}{3}\right) +\left( \dfrac{1}{3}-\dfrac{1}{4}%
\right) +...\left( \dfrac{1}{n}-\dfrac{1}{n+1}\right) =1-\dfrac{1}{n+1}
\end{eqnarray*}

In this case, each partial sum of the series has only a fixed number of
terms after cancellation. \ We call such a series a \textbf{telescoping sum}%
.\bigskip

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Definition: \ \textbf{Geometric series} are of the form 
\begin{equation*}
a+ar+ar^{2}+...+ar^{n-1}+....=\dsum\limits_{n=1}^{\infty }ar^{n-1}
\end{equation*}%
where $a$ and $r$ are fixed real numbers and $a\not=0$.%
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\medskip

\bigskip

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Theorem: \ If $\left\vert r\right\vert <1$, the geometric series $%
a+ar+ar^{2}+...+ar^{n-1}+...$ converges to $\dfrac{a}{1-r}$ and if $%
\left\vert r\right\vert \geq 1$, the series diverges.%
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\medskip

\bigskip

Example 2) \ Determine whether the given series converges or diverges. \ If
it converges, find the sum of the series.\bigskip

a) \ $\dsum\limits_{n=0}^{\infty }2^{n-1}\cdot 3^{2-n}$\bigskip

We start by re-writing $a_{n}$%
\begin{equation*}
a_{n}=2^{n-1}\cdot 3^{2-n}=\dfrac{2^{n}}{2}\cdot \dfrac{9}{3^{n}}=\dfrac{9}{2%
}\left( \dfrac{2}{3}\right) ^{n}
\end{equation*}%
We can now determine that $a=\dfrac{9}{2}$ and $r=\dfrac{2}{3}$. \ Since $%
-1<r<1$, the series converges and its sum is 
\begin{equation*}
s=\dfrac{a}{1-r}=\dfrac{~\dfrac{9}{2}~}{1-\dfrac{2}{3}}=\dfrac{27}{2}
\end{equation*}%
b) \ $\dsum\limits_{n=3}^{\infty }3^{2n-1}\cdot \left( -5\right) ^{2-n}$%
\bigskip

We start by re-writing $a_{n}$%
\begin{equation*}
a_{n}=3^{2n-1}\cdot \left( -5\right) ^{2-n}=\dfrac{9^{n}}{3}\cdot \dfrac{25}{%
\left( -5\right) ^{n}}=\dfrac{25}{3}\left( -\dfrac{9}{5}\right) ^{n}
\end{equation*}%
We can now determine that $a=\dfrac{25}{3}$ and $r=-\dfrac{9}{5}$. \ Since $%
r<-1$, the series diverges and so the sum is not defined. \ \ There are more
examples in the separate handout dedicated to \href{http://www.teaching.martahidegkuti.com/shared/lnotes/6_calculus/series/geometric_series/geometric_series.pdf%
}{geometric series}.\bigskip

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Theorem: \ If $\dsum\limits_{n=1}^{\infty }a_{n}$ converges, then $%
a_{n}\rightarrow 0$. 
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\medskip

\bigskip

Proof: \ Let $\dsum\limits_{n=1}^{\infty }a_{n}$ be a convergent series with
sum $s\in 
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$. \ Then $\lim\limits_{n\rightarrow \infty }s_{n}=s$. \ Clearly $%
a_{n}=s_{n}-s_{n-1}$.%
\begin{equation*}
\lim\limits_{n\rightarrow \infty }a_{n}=\lim\limits_{n\rightarrow \infty
}\left( s_{n}-s_{n-1}\right) =\lim\limits_{n\rightarrow \infty
}s_{n}-\lim\limits_{n\rightarrow \infty }s_{n-1}=s-s=0
\end{equation*}%
\bigskip

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Theorem: \ ($n$th term test)\ If $\lim\limits_{n\rightarrow \infty }a_{n}$
doesn't exist or exists but is not zero, then $\dsum\limits_{n=1}^{\infty
}a_{n}$ diverges. 
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\medskip

\bigskip

The "smallness" of the $n$th term is a necessary but not sufficient
condition of the convergence of the series. \ For example, $%
\dsum\limits_{n=1}^{\infty }\dfrac{1}{n}$ diverges even though the $n$th
term, $a_{n}$ approaches zero. \ The logical conclusion is that the $n$th
term test can only be used to prove divergence of a series.\bigskip

Example 3) \ Prove that $\dsum\limits_{n=1}^{\infty }\dfrac{n-2}{2n+1}$
diverges.\bigskip

Proof: \ $\lim\limits_{n\rightarrow \infty }a_{n}=\lim\limits_{n\rightarrow
\infty }\dfrac{n-2}{2n+1}=\lim\limits_{n\rightarrow \infty }\dfrac{1-\dfrac{2%
}{n}}{2+\dfrac{1}{n}}=\dfrac{1}{2}$. \ Since the $n$th term does not
approach zero, the series diverges.\bigskip

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Theorem: \ If $\dsum a_{n}=A$ \bigskip and $\dsum b_{n}=B$ are convergent
series, then

\qquad 1) \ Sum Rule \ \ $\dsum \left( a_{n}+b_{n}\right) =\dsum a_{n}+\dsum
b_{n}=A+B\medskip $

\qquad 2) \ Difference Rule \ \ \ \ $\dsum \left( a_{n}-b_{n}\right) =\dsum
a_{n}-\dsum b_{n}=A-B\medskip $

\qquad 3) \ Constant Multiple Rule: \ $\dsum ka_{n}=k\dsum a_{n}=kA$ \ \ \ \
for any $k\in 
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\medskip

\bigskip

These properties can be proved using the properties of limits of sequences.
Consequences: \ \bigskip

\qquad 1) \ Every non-zero constant multiple of a divergent series diverges.$%
\medskip $

\qquad 2) \ If $\dsum a_{n}$ converges and $\dsum b_{n}$ diverges, then $%
\dsum \left( a_{n}+b_{n}\right) $ and $\dsum \left( a_{n}-b_{n}\right) $
both diverge.\bigskip

Example 4.) \ Compute the sum of each of the following series or state if
the series diverges.\bigskip

a) \ $\dsum\limits_{n=0}^{\infty }\dfrac{7\cdot 2^{n}-3^{n}}{5^{n}}$\bigskip

Solution: \ 
\begin{equation*}
\dsum\limits_{n=0}^{\infty }\dfrac{7\cdot 2^{n}-3^{n}}{5^{n}}%
=\dsum\limits_{n=0}^{\infty }\left( \dfrac{7\cdot 2^{n}}{5^{n}}-\dfrac{3^{n}%
}{5^{n}}\right) =\dsum\limits_{n=0}^{\infty }7\left( \dfrac{2^{n}}{5^{n}}%
\right) -\dsum\limits_{n=0}^{\infty }\left( \dfrac{3^{n}}{5^{n}}\right)
=\dsum\limits_{n=0}^{\infty }7\left( \dfrac{2}{5}\right)
^{n}-\dsum\limits_{n=0}^{\infty }\left( \dfrac{3}{5}\right) ^{n}
\end{equation*}

We separately compute the sums of the geometric series. \ In the first
series, $a=7$ and $r=\dfrac{2}{5}$ and in the second, $a=1$ and $r=\dfrac{3}{%
5}$. \ 
\begin{equation*}
s_{1}=\dfrac{7}{1-\dfrac{2}{5}}=\dfrac{35}{3}\text{ \ and \ }s_{2}=\dfrac{1}{%
1-\dfrac{3}{5}}=\dfrac{5}{2}
\end{equation*}%
and so the difference of the two series is $\dfrac{35}{3}-\dfrac{5}{2}=%
\dfrac{55}{6}$.\bigskip

b) \ $\dsum\limits_{n=0}^{\infty }e^{n}+e^{-n}$\bigskip

Solution: \ 
\begin{equation*}
\dsum\limits_{n=0}^{\infty }e^{n}+e^{-n}=\dsum\limits_{n=0}^{\infty
}e^{n}+\dsum\limits_{n=0}^{\infty }e^{-n}
\end{equation*}%
\bigskip

The first series diverges because it is a geometric series with $r=e>1$. \
Thus the sum of the two series diverges as well.\bigskip

Recall the Monotonic Sequence Theorem: \ If a non-decreasing sequence is
bounded from above, then it is convergent. \ A consequence of this is the
following theorem.\bigskip

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Theorem: \ A series $\dsum\limits_{n=1}^{\infty }a_{n}$ of non-negative
terms converges if and only if its partial sums are bounded from above.%
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\medskip

\bigskip

Proof: \ If the sequence $\left\{ a_{n}\right\} $ has only non-negative
terms, then the sequence $\left\{ s_{n}\right\} $ of partial sums is
non-decreasing. \ \bigskip

Example 5.) \ The series $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n}$ diverges.

Proof: \ We group the terms into clusters that each add up to a number
greater than $\dfrac{1}{2}$. \ Since there are infinitely many such
clusters, the sequence of partial sums is not bounded from above. \ Thus the
series diverges. \ 
\begin{equation*}
1+\dfrac{1}{2}+\underset{>\dfrac{2}{4}}{\underbrace{\left( \dfrac{1}{3}+%
\dfrac{1}{4}\right) }}+\underset{>\dfrac{4}{8}}{\underbrace{\left( \dfrac{1}{%
5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right) }}+\underset{>\dfrac{8}{16}}%
{\underbrace{\left( \dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right) }}%
+....
\end{equation*}

This is an excellent example to demonstrate that the $n$th term may approach
zero and yet the series diverges.\bigskip

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Theorem: \ (The Integral Test) \ Let $\left\{ a_{n}\right\} $ be a sequence
of positive terms. \ Suppose that $a_{n}=f\left( n\right) $, where $f$ is a
continuous, positive, decreasing function of $x$ for all $x\geq N$ ($N$ a
positive integer). \ Then the series $\dsum\limits_{n=N}^{\infty }a_{n}$ and
the integral $\dint\limits_{N}^{\infty }f\left( x\right) dx$ both converge
or both diverge.%
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\medskip

\bigskip

Proof. \ Suppose that the conditions hold. \ Suppose that $N=1$. \ If the
function $f$ is decreasing, then all left-hand Riemann sums overestimate the
area under the graph and all right-hand sums underestimate the same area. \
Consider now the following:\bigskip

$a_{1}+a_{2}+a_{3}+...+a_{n}$ \ is a left sum for $f$ on the interval $\left[
1,n+1\right] $ with the partition $\left\{ 1,2,3,...,n+1\right\} $%
\begin{equation*}
\dint\limits_{1}^{n+1}f\left( x\right) dx\leq a_{1}+a_{2}+a_{3}+...+a_{n}
\end{equation*}

$a_{2}+a_{3}+...+a_{n}$ \ is a right sum for $f$ on $\left[ 1,n\right] $
with the partition $\left\{ 1,2,3,...,n\right\} $. \ Thus%
\begin{equation*}
a_{2}+a_{3}+...+a_{n}\leq \dint\limits_{1}^{n}f\left( x\right) dx
\end{equation*}%
Add $a_{1}$ to both sides:%
\begin{equation*}
a_{1}+a_{2}+a_{3}+...+a_{n}\leq a_{1}+\dint\limits_{1}^{n}f\left( x\right) dx
\end{equation*}%
And so we have that 
\begin{equation*}
\dint\limits_{1}^{n+1}f\left( x\right) dx\leq
a_{1}+a_{2}+a_{3}+...+a_{n}\leq a_{1}+\dint\limits_{1}^{n}f\left( x\right) dx
\end{equation*}%
These inequalities are true for all $n$, and continue to hold as $n$
approaches infinity.\bigskip

If $\dint\limits_{1}^{\infty }f\left( x\right) dx$ is finite, then $%
a_{1}+\dint\limits_{1}^{\infty }f\left( x\right) dx$ is an upper limit of
the sequence of partial sums, and so $\dsum\limits_{n=1}^{\infty }a_{n}$ is
finite. \ If $\dint\limits_{1}^{\infty }f\left( x\right) dx$ is infinite,
then $\dint\limits_{1}^{n+1}f\left( x\right) dx$ is not bounded from above
and so the sequence of partial sums is also not bounded from above, thus $%
\dsum\limits_{n=1}^{\infty }a_{n}$ is infinite. \ Thus the series and the
integral are both finite or both infinite.\bigskip

Example 6.) \ Determine whether the given series is convergent or
not.\bigskip

a) \ $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n}$\bigskip

This will be the second time we prove that this series diverges, but it is a
very famous series. \ It is called the harmonic series. \ We will use the
integral test. \ The sequence $1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},....$
\ is decreasing and all terms are positive. \ Also, the function $f\left(
x\right) =\dfrac{1}{x}$ is continuous, positive, and decreasing on $\left[
1,\infty \right) $, so we may apply the integral test. \ Since $%
\dint\limits_{1}^{\infty }\dfrac{1}{x}dx$ diverges, so does $%
\dsum\limits_{n=1}^{\infty }\dfrac{1}{n}$.\bigskip

b) \ $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$\bigskip

We will use the integral test. \ The sequence $1,\dfrac{1}{4},\dfrac{1}{9},%
\dfrac{1}{16},....$ \ is decreasing and all terms are positive. \ Also, the
function $f\left( x\right) =\dfrac{1}{x^{2}}$ is continuous, positive, and
decreasing on $\left[ 1,\infty \right) $, so we may apply the integral test.
\ Since $\dint\limits_{1}^{\infty }\dfrac{1}{x^{2}}dx$ converges, so does $%
\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$.\bigskip

c) \ $\dsum\limits_{n=3}^{\infty }\dfrac{1}{n\ln n}$\bigskip

We will use the integral test. \ The sequence $\dfrac{1}{3\ln 3},\dfrac{1}{%
4\ln 4},\dfrac{1}{5\ln 5},....$ \ is decreasing and all terms are positive.
\ Also, the function $f\left( x\right) =\dfrac{1}{x\ln x}$ is continuous, \
and positive on $\left[ 3,\infty \right) $. \ We may apply the integral
test, but first we must prove that $f\left( x\right) =\dfrac{1}{x\ln x}$ is
decreasing on $\left[ 3,\infty \right) $. \ We differentiate $f\left(
x\right) $%
\begin{equation*}
\dfrac{d}{dx}\left( \dfrac{1}{x\ln x}\right) =\dfrac{d}{dx}\left( \left(
x\ln x\right) ^{-1}\right) =-1\left( x\ln x\right) ^{-2}\left( \ln
x+1\right) =-\dfrac{1+\ln x}{\left( x\ln x\right) ^{2}}
\end{equation*}%
Since the derivative of $f$ is negative on $\left[ 3,\infty \right) $, the
function is decreasing and so the conditions for applying the integral test
all hold. \ We compute the improper integral $\dint\limits_{3}^{\infty }%
\dfrac{1}{x\ln x}dx$. \ We use integration by substitution to compute the
indefinite integral: \ let $u=\ln x$ then $du=\dfrac{1}{x}dx$%
\begin{equation*}
\dint \dfrac{1}{x\ln x}dx=\dint \dfrac{1}{\ln x}\left( \dfrac{1}{x}dx\right)
=\dint \dfrac{1}{u}du=\ln \left\vert u\right\vert +C=\ln \left\vert \ln
x\right\vert +C
\end{equation*}%
Now for the improper integral: \ 
\begin{equation*}
\dint\limits_{3}^{\infty }\dfrac{1}{x\ln x}dx=\lim\limits_{N\rightarrow
\infty }\dint\limits_{3}^{N}\dfrac{1}{x\ln x}dx=\lim\limits_{N\rightarrow
\infty }\ln \left\vert \ln x\right\vert \biggr\rvert_{3}^{N}=\lim\limits_{N%
\rightarrow \infty }\left( \ln \ln N-\ln \ln 3\right) =\infty
\end{equation*}

Since the integral diverges, so does the series $\dsum\limits_{n=3}^{\infty }%
\dfrac{1}{n\ln n}$.\bigskip \bigskip \bigskip \pagebreak

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

Determine which of the following series converge and which diverge.\bigskip

Please note that most of these problems can be solved using several
different methods.\bigskip 
%TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}%
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\begin{multicols}{3}%
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\begin{enumerate}
\item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}+2n}$ $\medskip $

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

\item $\dsum\limits_{n=0}^{\infty }\dfrac{1}{n^{2}+1}$ $\medskip $

\item $\dsum\limits_{n=0}^{\infty }\dfrac{5^{n}+3^{n}}{4^{n}}$ $\medskip $\ 

\item $\dsum\limits_{n=0}^{\infty }\sqrt{n}$ \ $\medskip $

\item $\dsum\limits_{n=1}^{\infty }\sqrt[n]{n}$ \ \ $\medskip $\ 

\item $\dsum\limits_{n=1}^{\infty }\dfrac{8}{n^{2}}$ \ \ $\medskip $\ 

\item $\dsum\limits_{n=2}^{\infty }\dfrac{\ln n}{n}\medskip $

\item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{1+n^{2}}\medskip $

\item $\dsum\limits_{n=1}^{\infty }\dfrac{2}{e^{n}}\medskip $

\item $\dsum\limits_{n=1}^{\infty }\func{sech}n$ \ $\medskip $

\item $\dsum\limits_{n=2}^{\infty }\dfrac{n+2}{n^{2}-n}\medskip $

\item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{3}}$ $\medskip $

\item $\dsum\limits_{n=2}^{\infty }\dfrac{1}{n\ln n}$ \ $\medskip $
\end{enumerate}

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{\LARGE \bigskip }

\pagebreak

\begin{center}
{\LARGE Answers - Practice Problems\bigskip }
\end{center}

1.) converges to $\dfrac{3}{4}\qquad $2.) \ converges to $1$ \qquad 3.)
converges \qquad 4.) $\ $\ diverges\qquad 5.) \ diverges\qquad 6.) \
diverges\bigskip

7.) \ converges\ \ \ \qquad 8.) \ diverges\qquad 9.) \ converges \qquad 10.)
\ converges \qquad 11.) \ converges\qquad 12.) \ diverges\bigskip

13.) \ converges\qquad 14.) \ diverges\bigskip \bigskip

\begin{center}
{\LARGE Solutions}

{\LARGE \bigskip }
\end{center}

\begin{enumerate}
\item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}+2n}$ converges to $\dfrac{%
3}{4}$, it is a telescoping sum $\ $

$a_{n}=\dfrac{1}{n^{2}+2n}=\dfrac{1}{n\left( n+2\right) }=\dfrac{\tfrac{1}{2}%
}{n}-\dfrac{\tfrac{1}{2}}{n+2}=\dfrac{1}{2}\left( \dfrac{1}{n}-\dfrac{1}{n+2}%
\right) $ \ \ \ \ \ \ \ 
\begin{eqnarray*}
s_{1} &=&a_{1}=\dfrac{1}{2}\left( \dfrac{1}{1}-\dfrac{1}{3}\right) =\dfrac{1%
}{3} \\
s_{2} &=&a_{1}+a_{2}=\dfrac{1}{2}\left[ \left( \dfrac{1}{1}-\dfrac{1}{3}%
\right) +\left( \dfrac{1}{2}-\dfrac{1}{4}\right) \right] =\dfrac{11}{24} \\
s_{3} &=&a_{1}+a_{2}+a_{3}=\dfrac{1}{2}\left[ \left( \dfrac{1}{1}-\dfrac{1}{3%
}\right) +\left( \dfrac{1}{2}-\dfrac{1}{4}\right) +\left( \dfrac{1}{3}-%
\dfrac{1}{5}\right) \right] =\dfrac{1}{2}\left( 1+\dfrac{1}{2}-\dfrac{1}{4}-%
\dfrac{1}{5}\right) =\dfrac{21}{40} \\
s_{3} &=&a_{1}+a_{2}+a_{3}=\dfrac{1}{2}\left[ \left( \dfrac{1}{1}-\dfrac{1}{3%
}\right) +\left( \dfrac{1}{2}-\dfrac{1}{4}\right) +\left( \dfrac{1}{3}-%
\dfrac{1}{5}\right) +\left( \dfrac{1}{4}-\dfrac{1}{6}\right) \right] =\dfrac{%
1}{2}\left( 1+\dfrac{1}{2}-\dfrac{1}{5}-\dfrac{1}{6}\right) =\dfrac{17}{30}
\\
s_{n} &=&a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} \\
&=&\dfrac{1}{2}\left[ \left( \dfrac{1}{1}-\dfrac{1}{3}\right) +\left( \dfrac{%
1}{2}-\dfrac{1}{4}\right) +\left( \dfrac{1}{3}-\dfrac{1}{5}\right)
+.....+\left( \dfrac{1}{n-1}-\dfrac{1}{n+1}\right) +\left( \dfrac{1}{n}-%
\dfrac{1}{n+2}\right) \right] \\
&=&\dfrac{1}{2}\left( 1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right) =%
\dfrac{1}{2}\left( \dfrac{3}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)
\end{eqnarray*}%
Thus $s_{n}=\dfrac{1}{2}\left( \dfrac{3}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}%
\right) $ and so 
\begin{equation*}
s=\lim\limits_{n\rightarrow \infty }s_{n}=\lim\limits_{n\rightarrow \infty }%
\dfrac{1}{2}\left( \dfrac{3}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right) =\dfrac{%
3}{4}
\end{equation*}

\item $\dsum\limits_{n=1}^{\infty }\dfrac{2n+1}{n^{2}\left( n+1\right) ^{2}}$
\ converges to $1$

Solution: \ We use partial fractions to decompose $a_{n}=\dfrac{2n+1}{%
n^{2}\left( n+1\right) ^{2}}$%
\begin{eqnarray*}
\dfrac{2n+1}{n^{2}\left( n+1\right) ^{2}} &=&\dfrac{A}{n}+\dfrac{B}{n^{2}}+%
\dfrac{C}{n+1}+\dfrac{D}{\left( n+1\right) ^{2}} \\
\dfrac{2n+1}{n^{2}\left( n+1\right) ^{2}} &=&\dfrac{An\left( n+1\right)
^{2}+B\left( n+1\right) ^{2}+Cn^{2}\left( n+1\right) +Dn^{2}}{n^{2}\left(
n+1\right) ^{2}} \\
2n+1 &=&An\left( n+1\right) ^{2}+B\left( n+1\right) ^{2}+Cn^{2}\left(
n+1\right) +Dn^{2}
\end{eqnarray*}%
Let $n=0$%
\begin{eqnarray*}
2n+1 &=&An\left( n+1\right) ^{2}+B\left( n+1\right) ^{2}+Cn^{2}\left(
n+1\right) +Dn^{2}\text{ \ \ becomes} \\
1 &=&B
\end{eqnarray*}%
Let $n=-1$%
\begin{eqnarray*}
2n+1 &=&An\left( n+1\right) ^{2}+B\left( n+1\right) ^{2}+Cn^{2}\left(
n+1\right) +Dn^{2}\text{ \ becomes} \\
-1 &=&D
\end{eqnarray*}%
Let $n=1$ 
\begin{eqnarray*}
2n+1 &=&An\left( n+1\right) ^{2}+B\left( n+1\right) ^{2}+Cn^{2}\left(
n+1\right) +Dn^{2}\text{ \ \ becomes} \\
3 &=&4A+4+2C-1 \\
0 &=&4A-2C \\
C &=&2A
\end{eqnarray*}%
Let $n=2$%
\begin{eqnarray*}
2n+1 &=&An\left( n+1\right) ^{2}+B\left( n+1\right) ^{2}+Cn^{2}\left(
n+1\right) +Dn^{2}\text{ \ \ becomes} \\
5 &=&18A+9+12C-4 \\
0 &=&18A+12C \\
0 &=&3A+2C
\end{eqnarray*}%
So we finally have this system: 
\begin{eqnarray*}
C &=&2A \\
3A+2C &=&0
\end{eqnarray*}%
we solve the system and obtain $A=C=0$. \ Thus the decomposition of $a_{n}$
is as follows:%
\begin{equation*}
a_{n}=\dfrac{2n+1}{n^{2}\left( n+1\right) ^{2}}=\dfrac{1}{n^{2}}-\dfrac{1}{%
\left( n+1\right) ^{2}}
\end{equation*}%
Thus the sequence of partial sums is as follows:%
\begin{eqnarray*}
s_{1} &=&a_{1}=\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}=1-\dfrac{1}{4}=\dfrac{3}{4}
\\
s_{2} &=&a_{1}+a_{2}=\left( \dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right) +\left( 
\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right) =1-\dfrac{1}{9}=\dfrac{8}{9} \\
s_{3} &=&a_{1}+a_{2}+a_{3}=\left( \dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right)
+\left( \dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right) +\left( \dfrac{1}{3^{2}}-%
\dfrac{1}{4^{2}}\right) =1-\dfrac{1}{16}=\dfrac{15}{16} \\
s_{n} &=&a_{1}+a_{2}+a_{3}+...+a_{n-1}+a_{n} \\
&=&\left( \dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right) +\left( \dfrac{1}{2^{2}}-%
\dfrac{1}{3^{2}}\right) +\left( \dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\right)
+...+\left( \dfrac{1}{\left( n-1\right) ^{2}}-\dfrac{1}{n^{2}}\right)
+\left( \dfrac{1}{n^{2}}-\dfrac{1}{\left( n+1\right) ^{2}}\right) \\
&=&1-\dfrac{1}{\left( n+1\right) ^{2}}
\end{eqnarray*}

\item $\dsum\limits_{n=0}^{\infty }\dfrac{1}{n^{2}+1}$ converges by integral
test

\item $\dsum\limits_{n=0}^{\infty }\dfrac{5^{n}+3^{n}}{4^{n}}$ \ \ \
diverges because 
\begin{equation*}
\dsum\limits_{n=0}^{\infty }\dfrac{5^{n}+3^{n}}{4^{n}}=\dsum\limits_{n=0}^{%
\infty }\dfrac{5^{n}}{4^{n}}+\dfrac{3^{n}}{4^{n}}=\dsum\limits_{n=0}^{\infty
}\left( \dfrac{5}{4}\right) ^{n}+\dsum\limits_{n=0}^{\infty }\left( \dfrac{3%
}{4}\right) ^{n}
\end{equation*}%
and the first geometric series diverges

\item $\dsum\limits_{n=0}^{\infty }\sqrt{n}$ \ \ \ diverges since $a_{n}$
does not approach $0$

\item $\dsum\limits_{n=1}^{\infty }\sqrt[n]{n}$ \ \ \ diverges since $a_{n}$
does not approach $0$

\item $\dsum\limits_{n=1}^{\infty }\dfrac{8}{n^{2}}$ \ \ \ converges since
it is a constant times $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$\ 

\item $\dsum\limits_{n=2}^{\infty }\dfrac{\ln n}{n}$ \ \ diverges by the
integral test: \ 

The function $f\left( x\right) =\dfrac{\ln x}{x}$ is continuous and positive
on \ $\left[ 2,\infty \right) $. \ \ We may apply the integral test if we
also show that $f$ is decreasing there. \ $\dfrac{d}{dx}\left( \dfrac{\ln x}{%
x}\right) =\dfrac{1}{x^{2}}\left( 1-\ln x\right) $ which is negative on $%
\left[ 3,\infty \right) $. \ So after $n=3,$ the conditions hold. \ $\dint 
\dfrac{\ln x}{x}dx=\dfrac{1}{2}\left( \ln x\right) ^{2}+C$. \ The improper
integral $\dint\limits_{3}^{\infty }\dfrac{\ln x}{x}dx$ diverges and
therefore so does the series.

\item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{1+n^{2}}$ \ \ converges by the
integral test

\item $\dsum\limits_{n=1}^{\infty }\dfrac{2}{e^{n}}$ \ \ \ \ \ \ \ \
converges, a geometric sequence with $a=\dfrac{2}{e}$ and $r=\dfrac{1}{e}$
with \ $0<\dfrac{1}{e}<1$

\item $\dsum\limits_{n=1}^{\infty }\func{sech}n$ \ \ \ \ \ \ \ \ \ \ $\ \ \ $%
converges by the integral test

$\dfrac{d}{dx}\left( \dfrac{2}{e^{x}+e^{-x}}\right) =2\dfrac{d}{dx}\left(
e^{x}+e^{-x}\right) ^{-1}=2\left( -1\right) \left( e^{x}+e^{-x}\right)
^{-2}\left( e^{x}-e^{-x}\right) $

$\dint \dfrac{2}{e^{x}+e^{-x}}dx=2\arctan \left( e^{x}\right) +C$ \ \ \ \ \ $%
\ \ \ \dint\limits_{0}^{\infty }\dfrac{2}{e^{x}+e^{-x}}dx=\dfrac{\pi }{2}$

\item $\dsum\limits_{n=2}^{\infty }\dfrac{n+2}{n^{2}-n}$ \ \ \ diverges

We decompose $a_{n}$ using partial fractions:%
\begin{equation*}
\dfrac{n+2}{n^{2}-n}=\dfrac{3}{n-1}-\dfrac{2}{n}
\end{equation*}%
Then $s_{n}$ becomes $s_{n}=2+\left( \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}%
+....\right) $

\begin{eqnarray*}
s_{n} &=&\dfrac{3}{1}\underbrace{-\dfrac{2}{2}+\dfrac{3}{2}}~\underbrace{-%
\dfrac{2}{3}+\dfrac{3}{3}}~\underbrace{-\dfrac{2}{4}+\dfrac{3}{4}}~-\dfrac{2%
}{5}+.....\underbrace{-\dfrac{2}{n-1}+\dfrac{3}{n-1}}~-\dfrac{2}{n} \\
&=&3+\left( \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...\dfrac{1}{n-1}\right) -%
\dfrac{2}{n}
\end{eqnarray*}%
This shows that the sequece of partial sums is not bounded from above, thus
the series diverges.

\item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{3}}$ converges by the
integral test

\item $\dsum\limits_{n=2}^{\infty }\dfrac{1}{n\ln n}$ \ \ \ \ diverges by
the integral test
\end{enumerate}

\vspace{1.2in}

\bigskip

\bigskip

\bigskip

\bigskip

\vspace{5in}

{\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
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