Summation and Product
$\sum_{i=1}^{n} a_n$
\sum_{i=1}^{n} a_n
\sum comes from sum. To write expressions below and above the sigma, use "_" and "^".
$\displaystyle \sum_{i=1}^n a_n$
\displaystyle \sum_{i=1}^n a_n
If you use displaystyle, the sigma will be larger. The formula will be placed above and below the sigma.
$\begin{eqnarray} \sum_{ k = 1 }^{ n } k^2 = \overbrace{ 1^2 + 2^2 + \cdots + n^2 }^{ n } = \frac{ 1 }{ 6 } n ( n + 1 ) ( 2n + 1 ) \end{eqnarray}$
\begin{eqnarray} \sum_{ k = 1 }^{ n } k^2 = \overbrace{ 1^2 + 2^2 + \cdots + n^2 }^{ n } = \frac{ 1 }{ 6 } n ( n + 1 ) ( 2n + 1 ) \end{eqnarray}
Using overbrace, you can display a brace on the top of the formula, and if you set the superscript, you can write text on top of the brace.
$\prod_{ i = 0 }^n x_i$
\prod_{ i = 0 }^n x_i
\prod comes from product.
$\displaystyle \prod_{i=0}^n x_i$
\displaystyle \prod_{i=0}^n x_i
$\begin{eqnarray} n! = \prod_{ k = 1 }^n k \end{eqnarray}$
\begin{eqnarray} n! = \prod_{ k = 1 }^n k \end{eqnarray}
This is a sample using factorial.
$\begin{eqnarray} \zeta (s) = \prod_{ p:\mathrm{ prime } } \frac{ 1 }{ 1-p^{-s} } \end{eqnarray}$
\begin{eqnarray} \zeta (s) = \prod_{ p:\mathrm{ prime } } \frac{ 1 }{ 1-p^{-s} } \end{eqnarray}
This is a sample using the Riemann zeta function.