Finding sum of a finite series. Consider the series 
$$\frac {q_1} {p_1} + \frac {q_1 q_2} {p_1 p_2} + \cdots + \frac  {q_1q_2 \cdots q_n} {p_1 p_2 \cdots p_n}$$ where $p_i + q_i = 1$ and $0 < p_i < 1$ and $0 <  q_i < 1$  for all $i=1,2, \cdots , n$.
How can I find the sum of this series? Please help me in this regard.
Thank you very much.
 A: Hint:
\begin{align}
  \frac{q_1}{p_1}+\frac{q_1 q_2}{p_1 p_2} &=
  \frac{q_1}{p_1} \left( 1+\frac{q_2}{p_2} \right) \\
  &= \frac{q_1}{p_1 p_2}
\end{align}
Updates for further thoughts:

The sum is not as trivial as we think at first glance.  First of all, we write the sum into Horner's form
$$S_n=\frac{q_1}{p_1}
\left \{
  1+\frac{q_2}{p_1}
  \left[
    1+\ldots+\frac{q_{n-1}}{p_{n-1}}
    \left( 1+\frac{q_n}{p_n} \right)
  \right]    
\right \}$$
and the summary for the first few cases: 
  \begin{align}
    S_1 &= \frac{q_1}{p_1} \\
    S_2 &= \frac{q_1}{p_1 p_2} \\
    S_3 &= \frac{q_1(q_2+p_2 p_3)}{p_1 p_2 p_3} \\
    S_4 &= \frac{q_1(q_2 q_3+p_3 p_4)}{p_1 p_2 p_3 p_4} \\
    S_5 &= \frac{q_1[q_2(q_3 q_4+p_4 p_5)+p_2 p_3 p_4 p_5]}
                {p_1 p_2 p_3 p_4 p_5} \\
    S_6 &= \frac{q_1[q_2 q_3(q_4 q_5+p_5 p_6)+p_3 p_4 p_5 p_6]}
                {p_1 p_2 p_3 p_4 p_5 p_6} \\
    S_7 &= \frac{q_1 \{q_2 [q_3 q_4(q_5 q_6+p_6 p_7)+p_4 p_5 p_6 p_7]+
                 p_2 p_3 p_4 p_5 p_6 p_7 \}}
                {p_1 p_2 p_3 p_4 p_5 p_6} \\
\end{align}

A: let
$$
\frac {q_1} {p_1} + \frac {q_1 q_2} {p_1 p_2} + \cdots + \frac  {q_1q_2 \cdots q_n} {p_1 p_2 \cdots p_n}=\frac{a_n}{p_1p_2\cdots p_n}
$$
$\implies a_1=q_1$ and $a_np_{n+1}+q_1q_2\cdots q_{n+1}=a_{n+1}$.
