% This is lecture no. 4 of MH1101/: Calculus II.

\section{Harmonic Series}%
\label{sec:harmonic_series}

The \textbf{harmonic series} is the following series:
\[
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dotsb + \frac{1}{n} + \dotsb = \sum_{n=1}^{\infty} \frac{1}{n}
.\]
Note that $\lim_{n \to \infty}a_n = \lim_{n \to \infty}\frac{1}{n}=0$.
\begin{proof}
	Divergence of the Harmonic Series.
	\begin{note}
		The sequence of partial sums $\left\{ s_{n} \right\}$ has a subsequence of the form $\left\{ s_{2^{k}} \right\}$.
	\end{note}
	\noindent
	Taking a look at the subsequence:
	\begin{align*}
		s_1 & = 1\\
		s_2 & = 1 + \frac{1}{2}\\
		s_4 & = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) \geqslant 1 + \frac{1}{2} + \left(\frac{1}{4}+\frac{1}{4}\right)= 1+ 2\left(\frac{1}{2}\right)\\
		s_8 & = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right)\\
		    & \geqslant 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} +\frac{1}{8}\right) = 1+ 3\left(\frac{1}{2}\right)
	\end{align*}
	Principally, we wish to establish the following inequality:
	\[
		s_{2^{n}}\geqslant 1+n\left(\frac{1}{2}\right)
	.\]
	We have verified that the inequality holds true for $n=0,1,2$.
	Now, suppose the following statement is true:
	\[
		s_{2^{n}}\geqslant 1 + n\left(\frac{1}{2}\right),\ n\in\mathbb{Z}
	.\]
	Now, we consider the case $n+1$:
	\begin{align*}
		s_{2^{n+1}} & = \underbrace{1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{2^{n}}}_{s_{2^{n}}} + \frac{1}{2^{n}+1} + \frac{1}{2^{n}+2} + \dotsb + \frac{1}{2^{n+1}}\\
			    & \geqslant s_{2^{n}} + \underbrace{\left(\frac{1}{2^{n}+2^{n}} + \frac{1}{2^{n}+2^{n}} + \dotsb + \frac{1}{2^{n}+2^{n}}\right)}_{\text{contains $2^{n}$ terms}}\\
		\intertext{Since $\frac{2^{n}}{2^{n+1}}=\frac{1}{2}$:}
			    & \geqslant 1 + n\left(\frac{1}{2}\right) + \frac{1}{2}\\
			    & \geqslant 1 + (n+1)\frac{1}{2}
	\end{align*}
	Continuing on, we see that:
	\[
	\lim_{n \to \infty}s_{2^{n}}=\infty
	.\]
	Therefore:
	\[
	\lim_{n \to \infty}s_{n}=\infty
	.\]
	Making the series divergent.
\end{proof}

\section{Properties}%
\label{sec:properties}

If $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are convergent series, then so are $c \sum_{n=1}^{\infty} a_n$, $\sum_{n=1}^{\infty} (a_n+b_n)$ and $ \sum_{n=1}^{\infty} (a_n-b_n)$, where $c$ is any constant.
The following laws apply:
\begin{itemize}
	\item $\sum_{n=1}^{\infty} ca_n = c \sum_{n=1}^{\infty} a_n$.
	\item $\sum_{n=1}^{\infty} (a_n+b_n) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n$.
	\item $\sum_{n=1}^{\infty} (a_n-b_n) = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{\infty} b_n$.
\end{itemize}
Because of the properties for convergent series, we have the following:
\begin{itemize}
	\item Every non-zero constant multiple of a divergent series is divergent.
	\item If $\sum_{n=1}^{\infty} a_n$ converges and $\sum_{n=1}^{\infty} b_n$ diverges, then $\sum_{n=1}^{\infty} (a_n \pm b_n)$ diverges.
\end{itemize}

\section{Test for Divergence}%
\label{sec:test_for_divergence}

\begin{thrm}
	Necessary condition for series convergence.
	If the series:
	\[
	\sum_{n=1}^{\infty} a_n
	.\]
	is a convergent series, then:
	\[
	\lim_{n \to \infty}a_n = 0
	.\]
\end{thrm}
\begin{note}
	$\lim_{n \to \infty}a_n=0$ does not imply that the series converges (Harmonic Series).
\end{note}
\begin{proof}
	Note that:
	\[
	a_n = s_n - s_{n-1}
	.\]
	Suppose that:
	\[
	\sum_{n=1}^{\infty} a_n=s
	.\]
	for some number $s$.
	Then:
	\[
	\lim_{n \to \infty}s_n = \lim_{n \to \infty}s_{n-1} = s
	.\]
	Therefore, we have:
	\[
		\lim_{n \to \infty}a_n = \lim_{n \to \infty}(s_n - s_{n-1}) = s-s = 0
	.\]
\end{proof}
The above result is useful in its contrapositive form:
\begin{mdframed}
	If $\lim_{n \to \infty}a_n$ does not exist or does not equal to 0, then the series $\sum_{n=1}^{\infty} a_n$ diverges.
\end{mdframed}
