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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.

little o
  • 4,853

2 Answers2

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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}

Ng Chung Tak
  • 18,990
  • I will definitely accept your answer. But before that can you please elaborate your answer by adding one more step? Then I can do it on my own. I have still some doubts regarding this calculation. Would you please make it clear to me by adding at least one more step to your comment? That will be really very helpful for me. Thank you very much. – little o Nov 05 '18 at 19:41
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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 $a_np_{n+1}+q_1q_2\cdots q_{n+1}=a_{n+1}$.

Ng Chung Tak
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Samvel Safaryan
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