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\begin{example*}[Geometric Series]
    $x \in \RR$, set $a_n = x^{n - 1}$ for $n \ge
    1$. Now
    \[ s_n = \sum_{j = 1}^n a_j = 1 + x + x^2 +
    \cdots + x^{n - 1} \]
    Then
    \[ s_n = \begin{cases}
        \frac{1 - x^n}{1 - x} & \text{for $x \neq
        1$} \\
        n & \text{for $x = 1$}
    \end{cases} \]
    \[ xs_n = x + x^2 + \cdots + x^n = s_n - 1+
    x^n \]
    \[ s_n(1 - x) = 1 - x^n \]
    if $|x| < 1$, $x^n \to 0$ and $s_n \to
    \frac{1}{1 - x}$. If $x > 1$, $x^n \to \infty$
    and $s_n \to \infty$. (Note $s_n \to \infty$
    if given $A$, there exists $N$ such that $s_n
    > A$ such that $s_n > A \,\,\forall\,\, n \ge
    N$, and $s_n \to -\infty$ if given $A$ there
    exists $N$ such that $s_n < -A$ for all $n \ge
    N$.)
    If $x < -1$ then $s_n$ does not converge
    (oscillates). If $x = -1$ then
    \[ s_n = \begin{cases}
        1 & \text{$n$ odd} \\
        0 & \text{$n$ even}
    \end{cases} \]
    Thus the geometric series converges if and
    only if $|x| < 1$.
\end{example*}

\noindent
To see for example that $x^n \to 0$ if $|x| < 1$,
consider first the case $0 < x < 1$. Write
$\frac{1}{x} = 1 + \delta$, $\delta > 0$. So
\[ x^n = \frac{1}{(1 + \delta)^n} \le \frac{1}{1 +
\delta n} \to 0 .\]
because $(1 + \delta)^n \ge 1 + n\delta$ from
binomial expansion. An easy observation from this
is that:

\begin{lemma}
    If $\sum_{j = 1}^\infty a_n$ converges, then
    $\lim_{j \to \infty} a_j = 0$.
\end{lemma}

\begin{proof}
    \[ s_n = \sum_{j = 1}^n a_j \]
    Then
    \[ a_n = s_n - s_{n - 1} .\]
    If $s_n \to a$, then $a_n \to 0$ (since $s_{n
    - 1} \to a$ as well).
\end{proof}

\begin{remark*}
    The converse of lemma 1.7 is false! For
    example, $\sum_{j = 1}^\infty \frac{1}{j}$
    diverges (harmonic series).
    \[ s_n = \sum_{j = 1}^n \frac{1}{j} \]
    \[ s_{2n} = s_n + \frac{1}{n + 1} + \frac{1}{n
    + 2} + \cdots + \frac{1}{n + n} > s_n +
    \frac{1}{2} \]
    since $\frac{1}{n + k} \ge \frac{1}{2n}$ for
    $k = 1, 2, \dots, n$. So if $s_n \to a$, then
    $s_{2n} \to a$ also, and thus $a \ge a +
    \half$ \contradiction
\end{remark*}

\subsubsection*{Series of Non-negative Terms}

$a_n \ge 0$. Basic result:

\begin{theorem}[The comparison test]
    Suppose that $0 \le b_n \le a_n
    \,\,\forall\,\, n$. Then if $\sum_{j =
    1}^\infty a_j$ converges, then so does
    $\sum_{j = 1}^\infty b_j$.
\end{theorem}

\begin{proof}
    Let $s_n = \sum_{j = 1}^N a_j$ and let $d_N =
    \sum_{j = 1}^N b_j$. Since $b_n \le a_n$ we
    have that $d_N \le s_N$. But $s_N \to s$, so
    $d_N \le s_N \le s \,\,\forall\,\, N$. Also,
    $d_N$ is an increasing sequence bounded above,
    hence $d_N$ converges.
\end{proof}

\begin{example*}
    \[ \sum_{j = 1}^\infty \frac{1}{n^2} \]
    \[ \ub{\frac{1}{n^2} < \frac{1}{n(n - 1)}}_{n
    \ge 2} = \frac{1}{n - 1} - \frac{1}{n} = a_n \]
    So
    \[ \sum_{j = 2}^N a_n = 1 - \frac{1}{2} +
    \frac{1}{2} - \frac{1}{3} + \cdots +
    \frac{1}{N - 1} - \frac{1}{N} = 1 -
    \frac{1}{N} \to 1 \]
    So by comparison, $\sum_{j = 1}^N
    \frac{1}{n^2}$ converges. In fact we get that
    \[ \sum_{j = 1}^\infty \frac{1}{n^2} \le 1 + 1
    = 2 .\]
\end{example*}

\begin{theorem}[Root test / Cauchy's test for convergence]
    Assume $a_n \ge 0$ and $a_n^{1 / n} \to a$ as
    $n \to \infty$. Then if $a < 1$, $\sum a_n$
    converges; if $a > 1$, $\sum a_n$ diverges.
\end{theorem}

\begin{remark*}
    Nothing can be said if $a = 1$ (examples
    coming up).
\end{remark*}

\begin{proof}
    If $a < 1$, choose $a < r < 1$. By definition
    of limit and hypothesis, there exists $N$ such
    that for all $n \ge N$,
    \[ a_n^{1 / n} < r \implies a_n < r^n \]
    But since $r < 1$, the geometric series $\sum
    r^n$ converges, so by theorem 1.8, $\sum a_n$
    converges. If $a > 1$, then for $n \ge N$,
    then $a_n^{1 / n} > 1 \implies a_n > 1$, thus
    $\sum a_n$ diverges (since $a_n$ does
    \emph{not} tend to zero).
\end{proof}

\begin{theorem}[Ratio test / D'Alembert's test]
    Suppose $a_n > 0$ and $\frac{a_{n + 1}}{a_n}
    \to \ell$. If $\ell < 1$, $\sum a_n$
    converges. If $\ell > 1$, $\sum a_n$
    converges.
\end{theorem}

\begin{note*}
    As before, nothing can be said for $\ell = 1$.
\end{note*}

\begin{proof}
    Suppose $\ell < 1$ and choose $r$ with $\ell <
    r < 1$. Then there exists $N$ such that
    for all $n \ge N$,
    \[ \frac{a_{n + 1}}{a_n} < r \]
    Therefore
    \[ a_n = \frac{a_n}{a_{n - 1}} \frac{a_{n -
    1}}{a_{n - 2}} \cdots \frac{a_{N + 1}}{a_N}
    a_N < a_N r^{n - N} \]
    \[ \implies a_n < Kr^n \]
    with $K$ \emph{independent} of $n$. Since
    $\sum r^n$ converges, so does $\sum a_n$ by
    theorem 1.8. If $\ell > 1$, choose $1 < r <
    \ell$, then $\frac{a_{n + 1}}{a_n} > r$ for
    all $n \ge N$, and as before
    \[ a_n = \frac{a_n}{a_{n - 1}} \frac{a_{n -
    1}}{a_{n - 2}} \cdots \frac{a_{N + 1}}{a_N}
    a_N > a_N r^{n - N} > a_N \]
    so $\sum a_n$ diverges.
\end{proof}