%The following command print all the series and their expansion
%which are typeset in the first horizontal part of the page 10.
%
%The macro has one parameter.
%      1) The width to use to typeset the formulae
%
\newcommand\TTenSerieTwo[1]{%
   %This command typeset a line equation about the series on page 10
   %
   %The command has 4 parameters
   %      1) The first formula to be expanded
   %      2) its expanded version
   %      3) The second formula to be expanded
   %      4) its expanded version
   \def\LineOfArray##1##2##3##4{%
     ##1&=&##2&##3&=&##4\\[\TTenExpansionSkip]%
   }%
   \parbox[t]{#1}{%
      \TTenSeriesFontSize
      \TTenTitle{Expansions:}%
            $\displaystyle
             \begin{array}{l@{\hspace{.1em}}c@{\hspace{.2em}}ll@{\hspace{.1em}}c@{\hspace{.2em}}l}%
                  \LineOfArray{\frac{1}{(1-x)^{n+1}}\ln\frac{1}{1- x}}%
                              {\sum_{i=0}^\infty (H_{n+i} - H_n) \binom{n+i}{i} x^i}%
                              {\left(\frac{1}{x}\right)^{\overline{-n}}}%
                              {\sum_{i=0}^\infty \SousEnsemble{i}{n} x^i}%
                  \LineOfArray{x^{\overline{n}}}%
                              {\sum_{i=0}^\infty  \Cycle{n}{i} x^i}%
                              {(e^x - 1)^n}%
                              {\sum_{i=0}^\infty \SousEnsemble{i}{n} \frac{n! x^i}{i!}}%
                  \LineOfArray{\left(\ln \frac{1}{1 -x}\right)^n}%
                              {\sum_{i=0}^\infty  \Cycle{i}{n} \frac{n! x^i}{i!}}%
                              {x \cot x}%
                              {\sum_{i=0}^\infty \frac{(-4)^i B_{2i} x^{2i}}{(2i)!}}%
                  \LineOfArray{\tan x}%
                              {\sum_{i=1}^\infty (-1)^{i-1}\frac{2^{2i} (2^{2i} - 1) B_{2i} x^{2i-1}}{(2i)!}}%
                              {\zeta(x)}%
                              {\sum_{i=1}^\infty \frac{1}{i^x}}%
                  \LineOfArray{\frac{1}{\zeta(x)}}%
                              {\sum_{i=1}^\infty \frac{\mu(i)}{i^x}}%
                              {\frac{\zeta(x-1)}{\zeta(x)}}%
                              {\sum_{i=1}^\infty \frac{\phi(i)}{i^x}}%
          \end{array}$%
  }%
}