\documentclass[a4paper,12pt]{article}
\begin{document}

\noindent {\bf Question}

\noindent Prove that $\sum_{n=1}^\infty \frac{1}{n^s}$ diverges
for $s <1$, by estimating its partial sums.
\medskip

\noindent {\bf Answer}

\noindent Note that for $s <1$, we have that $n^s <n$ (even for $s
=0$ or $s$ negative), and hence that $\frac{1}{n^s} >
\frac{1}{n}$. Hence, if we let $S_k$ be the $k^{th}$ partial sum
of the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$, and $T_k$
be the $k^{th}$ partial sum of the series $\sum_{n=1}^\infty
\frac{1}{n^s}$ under consideration, then $T_k > S_k$.  Since
$\frac{1}{n^s} >0$ for all $n$, we have that $\{ T_k\}$ is an
unbounded monotonically increasing sequence, unbounded since $\{
S_k\}$ is unbounded by the argument given below, and so $\{ T_k\}$
diverges.  So, by definition, $\sum_{n=1}^\infty \frac{1}{n^s}$
diverges.

\medskip

\noindent Argument

\noindent The series $\sum_{n=1}^\infty \frac{1}{n^s}$ converges
if and only if $s >1$.

\medskip
\noindent For $s =1$, this series is called the {\bf harmonic
series}, and we can prove directly that it diverges.  Note that
$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$, that $\frac{1}{5} +
\cdots + \frac{1}{8} > 4\frac{1}{8} = \frac{1}{2}$, and in general
that
\[ \frac{1}{2^{k-1} +1} + \frac{1}{2^{k-1} +2} +\cdots +\frac{1}{2^k}
> 2^{k-1}\frac{1}{2^k} =\frac{1}{2}. \]
Hence, the $(2^k)^{th}$ partial sum $S_{2^k}$ satisfies $S_{2^k}
>1 +k\frac{1}{2}$.  Since the terms in the harmonic series are all
positive, the sequence of partial sums is monotonically
increasing, and by the calculation done the sequence of partial
sums is unbounded, and so the sequence of partial sums diverges.
Hence, the harmonic series diverges.
\end{document}