\problemdata                                                             
        {Riemann's Zeta Function Derivation} % latex-friendly label for the prob.
        {Summations} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S01 Tutorial 6, 4} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
Riemann's Zeta Function $\zeta(k)$ is defined to be the infinite summation:
\[
1+\frac1{2^k}+\frac1{3^k} \cdots = \sum_{j \ge 1} \frac1{j^k}
\]
Prove that 
\[
\sum_{k \ge 2} (\zeta(k)-1) = 1
\]

%\textbf{Hint:} Recall from problem set~5, problem 2, that
%$\sum_{k=2}^n {\frac{1}{k(k-1)}} = 1 - \frac{1}{n}$.
\solution{
 \textbf{Solution:}
 First, we prove that $\sum_{k=2}^n {\frac{1}{k(k-1)}} = 1 - \frac{1}{n}$
 By using partial fractions, we find that $\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$.\\
So we have:
\begin{eqnarray*}
\sum_{k=2}^n {\frac{1}{k(k-1)}} & = & 
\sum_{k=2}^n \left(\frac{1}{k-1} - \frac{1}{k} \right)  \\
 & = & \sum_{k=2}^n \frac{1}{k-1} - \sum_{k=2}^n\frac{1}{k}  \\
 & = & \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{n-2} + \frac{1}{n-1} \;\; -\\
 &   & \left[ \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n-1} + \frac{1}{n} \right]
\end{eqnarray*}

So all that does not cancel out is $1 - \frac{1}{n}$.

We now prove that $\sum_{k \ge 2} (\zeta(k)-1) = 1$ 
  \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] \\
    & = & \sum_{k \ge 2} \sum_{j \ge 2} \frac{1}{j^k} \\
    & = & \sum_{j \ge 2} \sum_{k \ge 2} \frac{1}{j^k} \\
    & = & \sum_{j \ge 2} \frac{1}{j^2} \sum_{k \geq 0} \frac{1}{j^k} \\
    & = & \sum_{j \ge 2} \frac{1}{j^2} \cdot \frac{1}{1 - 1/j} \\
    & = & \sum_{j \ge 2} \frac{1}{j (j - 1)} \\
    & = & \lim_{n \rightarrow \infty} \sum_{j = 2}^{n} \frac{1}{j (j - 1)} \\
    & = & \lim_{n \rightarrow \infty} (1 - \frac{1}{n}) \\
    & = & 1
  \end{eqnarray*}
  } 

\end{problem}
                                 