\problemdata      
        {?} % latex-friendly label for the prob.
        {Summations} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, TU7} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem}
Riemann's Zeta Function $\zeta(k)$ is defined to be the infinite
summation:
\[
1+\frac1{2^k}+\frac1{3^k} \cdots = \sum_{j \ge 1} \frac1{j^k}
\]
Below is a proof that 
\[
\sum_{k \ge 2} (\zeta(k)-1) = 1
\]
Justify each line of the proof.
\begin{eqnarray}
\sum_{k \ge 2} (\zeta(k)-1) %
& = & \sum_{k \ge 2} \left[ \left( \sum_{j \ge 1} \frac{1}{j^k} \right) - 1 \right] \label{eq:1} \\
& = & \sum_{k \ge 2} \sum_{j \ge 2} \frac{1}{j^k} \label{eq:2} \\
& = & \sum_{j \ge 2} \sum_{k \ge 2} \frac{1}{j^k} \label{eq:3} \\
& = & \sum_{j \ge 2} \frac{1}{j^2} \sum_{k \geq 0} \frac{1}{j^k} \label{eq:4} \\
& = & \sum_{j \ge 2} \frac{1}{j^2} \cdot \frac{1}{1 - 1/j} \label{eq:5} \\
& = & \sum_{j \ge 2} \frac{1}{j (j - 1)} \label{eq:6} \\
& = & \lim_{n \rightarrow \infty} \sum_{j = 2}^{n} \frac{1}{j (j - 1)} \label{eq:7} \\
& = & \lim_{n \rightarrow \infty} \sum_{j = 2}^{n} \frac{1}{j-1} -\frac{1}{j} \label{eq:8} \\
& = & \lim_{n \rightarrow \infty} (1 - \frac{1}{n}) \label{eq:9} \\
& = & 1 \label{eq:10} 
\end{eqnarray}

\solution{
\eqref{eq:1}~Definition of $\zeta(k)$.
\eqref{eq:2}~Because ${\displaystyle \sum_{j \ge 1} \cfrac{1}{j^k} = 1 + \sum_{j \ge 2} \cfrac{1}{j^k}}$.
\eqref{eq:3}~Reordering; this is ok because at this point all terms are positive.
\eqref{eq:4}~Because 
$\displaystyle
\sum_{k \ge 2} \cfrac{1}{j^k} = 
\sum_{k \ge 0} \cfrac{1}{j^{k+2}} = 
\sum_{k \ge 0} \cfrac{1}{j^k\cdot j^2} =
\cfrac{1}{j^2} \sum_{k \ge 0} \cfrac{1}{j^k}
$.
\eqref{eq:5}~Theorem~2.2 from Notes~\#7.
\eqref{eq:6}~Algebra inside every summand.
\eqref{eq:7}~Definition of infinite summation.
\eqref{eq:8}~Algebra inside every summand.
\eqref{eq:9}~The sum telescopes: 1 is added once; every one of 
$\cfrac{1}{2}, \cfrac{1}{3}, \dots, \cfrac{1}{n-1}$ is subtracted once
and then added once; $\cfrac{1}{n}$ is subtracted once.
\eqref{eq:10}~Simple limits.
}
\end{problem}