Finding sum of a finite series.

Question

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.

Answer 1

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}

Answer 2

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 $a_np_{n+1}+q_1q_2\cdots q_{n+1}=a_{n+1}$.