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\begin{center}

ANALYSIS

REAL VARIABLE

SEQUENCES

\end{center}

\begin{description}

\item[]
A monotonic increasing sequence either converges to a finite limit
or diverges to $=\infty$.

\item[Proof]
Let $S$ be the set of numbers in the sequence $\{a_n\}$.

\begin{description}

\item[Case (i)]
$S$ has no upper bound.

Given any $X,\exists n_0/\ a_{n_0}&gt;X\Rightarrow a_n&gt;X\textrm{ for
}n\geq n_0$.

This means that the sequence diverges to $+\infty$.

\item[Case (ii)]
$S$ has an upper bound;\ $\overline{bd}a_n=a$.

Given $\varepsilon&gt;0\exists n_1/\ a_{n_1}&gt;a-\varepsilon\Rightarrow
a-\varepsilon&lt;a_n\leq a$ for $n\geq n_1$.

This means that the sequence converges to $a$.

\end{description}

\item[Upper and lower limits]

\item[Definition]

$$\left.\begin{array}{l}\lim_{n\to\infty}sup\ a_n\\
\overline{\lim_{n\to\infty}}a_n\end{array}\right\}=\lim_{n\to\infty}
{sup}_{m\geq n}a_m$$

$$\left.\begin{array}{l}\lim_{n\to\infty}inf\ a_n\\
\underline{\lim}_{n\to\infty}a_n\end{array}\right\}=\lim_{n\to\infty}
{inf}_{m\geq n}a_m$$


\item[Justification (for lim sup)]

\begin{description}

\item[Case (i)]
Suppose $S$ has no upper bound.

i.e $sup_{m\geq 1}a_m=+\infty$

Then $sup_{m\geq n}a_m=+\infty$ for all $n$.

In these circumstances we say that
$\overline{\lim_{n\to\infty}}a_n=+\infty$

\item[Case (ii)]
Now suppose $S$ has an upper bound. Write $sup_{m\geq
n}a_m=A^{(n)}$

$A^{(1)}A^{(2)}\ldots$ is a monotonic decreasing sequence. If this
sequence is bounded below, then it converges to a number
$\Lambda$, and $\overline{\lim_{n\to\infty}}a_n=\Lambda$, and if
not, $\overline{\lim_{n\to\infty}}a_n=-\infty$

\end{description}

Subject to the conventions we have introduced $\overline{\lim}a_n$
always exists.

 Suppose $\overline{\lim}a_n=\Lambda$ and suppose
$\Lambda$ finite. Then $\Lambda$ has the following property (P)

\begin{description}

\item[(P)]
For every $\varepsilon&gt;0$

\item[(i)]
$a_n&gt;\Lambda-\varepsilon$ for an infinity of $n$

\item[(ii)]
$a_n&gt;\Lambda+\varepsilon$ for at most a finite number of $n$.

\end{description}

\item[Proof]

\begin{description}

\item[(i)]
If $\exists$ only a finite number of $n/\ a_n&gt;\Lambda-\varepsilon$
then $\Rightarrow\overline{\lim}a_n\leq\Lambda-\varepsilon$.

\item[(ii)]
If $\exists$ an infinity of $n/\
a_n&gt;\Lambda+\varepsilon\Rightarrow\overline{\lim}a_n\geq\Lambda+\varepsilon$

\end{description}
We may define $\Lambda$ as the upper bound of the set of limit
points of $\{a_n\}$.

\item[Theorem 1]
$$\underline{\lim}a_n\leq\overline{\lim}a_n$$

\item[Proof]
$$inf_{m\geq n}a_m\leq sup_{m\geq n}a_m$$

Let $n\to\infty$ and the result follows.

\item[Theorem 2]
$$\lim_{n\to\infty}a_n=a\Leftrightarrow\underline{\lim}_{n\to\infty}a_n
=\overline{\lim_{n\to\infty}}a_n=a$$

\item[Proof]

\begin{description}

\item[(1)]
Suppose that $\lim_{n\to\infty}a_n=a$

Then given $\varepsilon&gt;0,\exists N=N(\varepsilon)/\
|a_m-a|&lt;\varepsilon$ for $m\geq n$. Hence foe $n\geq N$

$$a-\varepsilon&lt;inf_{m\geq n}a_m\leq sup_{m\geq
n}a_m&lt;a+\varepsilon$$

Hence
$a-\varepsilon&lt;\lambda\leq\Lambda&lt;a+\varepsilon&lt;a+\varepsilon\Rightarrow\lambda=\Lambda$

\item[(2)]
Suppose
$\underline{\lim}_{n\to\infty}a_n=\overline{\lim_{n\to\infty}}a_n=a$

Then, given $\varepsilon&gt;0\ \exists N=N(\varepsilon)/\ $ for
$n\geq N,$

$$a-\varepsilon\leq inf_{m\geq n}a_m\leq sup_{m\geq n}a_m\leq
a+\varepsilon$$

Hence for $m\geq N$ we have

$$|a_m-a|\leq\varepsilon$$

\end{description}

\item[Theorem 3]
It is possible to choose a subsequence $b_1,b_1\ldots$ from
$(a_1,a_2\ldots,/\lim_{n\to\infty}b_n$ exists
$=\overline{\lim_{n\to\infty}}a_n$

\item[Proof]

\begin{description}

\item[(i)]
$\Lambda=+\infty$

\item[(ii)]
$\Lambda=-\infty$

\item[(iii)]
$\Lambda$ finite

Let $I_n$ be the interval
$[\Lambda-\frac{1}{n},\Lambda+\frac{1}{n}]$

&gt;From property (P);

\begin{tabular}{ll}
$\exists r_1/\ a_{r_1}\in I_1$&amp;Write $b_1=a_{r_1}$\\ $\exists
r_2&gt;r_1/\ a_{r_2}\in I_2$&amp;Write $b_2=a_{r_2}$
\end{tabular}

Continue the process, and we obtain a subsequence of the required
type.

\end{description}

\item[Corollary]
A bounded sequence of points in $n$-dimensional Euclidean space
contains a convergent subsequence (In particular the result is
true for sequences of real and complex numbers $[R_1\&amp;R_2]$

\item[Example]
Every real sequence contains either a strictly monotonic
increasing subsequence, or a weakly monotonic decreasing
subsequence.

\item[Theorem 4 The General (Cauchy) Principle of Convergence]
The following condition (c) is a necessary and sufficient
condition for $\lim_{n\to\infty}a_n$ to exist.

(c) Given $\varepsilon&gt;0\exists N=N(\varepsilon)/\
|a_m-a_n|&lt;\varepsilon$ whenever $m&gt;n\geq N$.

\item[Proof]

\begin{description}

\item[Necessity]
Suppose $\lim_{n\to\infty}a_n=a$

$\exists N/\ |a_s-a|&lt;\frac{1}{2}\varepsilon$ for $s\geq N$
therefore for $m&gt;n&gt;N$

$$|a_m-a_n|\leq|a_m-a|+|a-a_n|&lt;\varepsilon$$

\item[Sufficiency]

First Proof.

\begin{description}

\item[(i)]
We prove the sequence bounded.

For $m&gt;n\geq N(1)$ we have

$$|a_m-a_n|&lt;1$$

Choose a fixed $n_1\geq N(1)$. Then $|a_m|&lt;|a_{n_1}|+1$ for $m\geq
n_1.$ Therefore $\{a_m\}$ is bounded.

\item[(ii)]
We show that, given $\varepsilon&gt;0$,

$$\overline{\lim}a_n-\underline{\lim}a_n&lt;\varepsilon.$$

Choose $N=N(\varepsilon)/\ |a_m-a_n|&lt;\varepsilon$ for $m&gt;n\geq N$.

Then for any $n\geq N$

$$sup_{u\geq n} a_u-inf_{v\geq n}a_v&lt;\varepsilon$$

Letting $n\to\infty$ we have

$$\Lambda-\lambda\leq\varepsilon$$

therefore

$$\Lambda=\lambda$$

\end{description}

Second Proof

\begin{description}

\item[(i)]
The sequence is bounded as before.

\item[(ii)]
$\exists$ a convergent subsequence $\{a_{n_r}\}$ converging to
$a$, say.

\item[(iii)]
We now show that $\{a_n\}$ converges to $a$.

Given $\varepsilon&gt;0\exists N=M(\varepsilon)$ such that

\begin{tabular}{lll}
(I)&amp;$|a-a_{n_r}|&lt;\frac{1}{2}\varepsilon$&amp;for $n_r\geq M$\\
(II)&amp;$|a_m-a_n|&lt;\frac{1}{2}\varepsilon$&amp;for $m&gt;n\geq M$
\end{tabular}

Let $r^*$ be the least $r$ for which $n_{r^*}\geq M$.

Then for $m&gt;n_{r^*}$ we have

$$|a_m-a|\leq|a_m-a_{n_{r^*}}|+|a_{n_{r^*}}-a|&lt;\frac{1}{2}\varepsilon+\frac{1}{2}\varepsilon$$

Therefore the sequence is convergent.

The above theorem applies to complex numbers and to
$n$-dimensional Euclidean space.

\end{description}

\end{description}

\item[Theorem 5]
$\sum_{n=1}^\infty a_n$ is convergent $\Leftrightarrow$ (c) given
$\varepsilon&gt;0\exists N=N(\varepsilon)/\
\left|\sum_{r=n}^ma_r\right|&lt;\varepsilon$ whenever $m&gt;n\geq N$

\item[Proof]
Apply theorem 4 to the sequence of partial sums and the result
follows.

\end{description}

\end{document}
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