Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms

Question

Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms

I first multiplied and divided $S$ with $1\cdot3\cdot5$ $$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{1\cdot3\cdot5\cdot7\cdot9}+\frac{1\cdot3\cdot5}{1\cdot3\cdot5\cdot7\cdot9\cdot11}+\cdots$$ Using the expansion of $(2n)!$ $$1\cdot3\cdot5\cdots(2n-1)=\frac{(2n)!}{2^nn!}$$ $$S=15\left[\sum_{r=1}^{20}\frac{\frac{(2r)!}{2^rr!}}{\frac{(2(r+3))!}{2^{r+3}(r+3)!}}\right]$$ $$S=15\cdot8\cdot\left[\sum_{r=1}^{20}\frac{(2r)!}{r!}\cdot\frac{(r+3)!}{(2r+6)!}\right]$$ $$S=15\sum_{r=1}^{20}\frac{1}{(2r+5)(2r+3)(2r+1)}$$

How can I solve the above expression? Or is there an simpler/faster method?

Answer 1

Hint:

$\frac{1}{(2r+5)(2r+3)(2r+1)}=\frac{1}{4}\left(\frac{1}{(2r+3)(2r+1)}-\frac{1}{(2r+5)(2r+3)}\right)$

Answer 2

Here's another approach, using recurrence relations.

Note that $$\begin{align} a_1&=\frac 17\qquad\text{and}\\ a_n&=a_{n-1}\left(\frac{2n-1}{2n+5}\right)\\ (2n+1)a_n+4a_n&=(2n-1)a_{n-1}\\ a_n&=\frac 14\big(b_{n-1}-b_n\big)\\ \end{align}$$ where $b_m=(2m+1)a_m$.

Summing from $2$ to $n$ by telescoping and adding $a_1$: $$\begin{align} \sum_{i=1}^{n}a_i &=\frac 14(b_1-b_n)+a_1\\ &=\frac 14\big(1-(2n+1)a_n\big) \qquad\qquad\qquad\text{as $b_1=3a_1$ and $a_1=\frac 17$}\\ &=\frac 14\bigg(1-\color{green}{(2n+1)}\cdot\frac{1\cdot 3\cdot 5\cdot \color{blue}{7\cdot 9\cdots(2n-1)}}{\color{blue}{7\cdot 9\cdot 11\cdots (2n-1)}\color{green}{(2n+1)}(2n+3)(2n+5)}\bigg)\\ &=\frac 14\bigg(1-\frac{15}{(2n+3)(2n+5)}\bigg)\\ &=\frac{n(n+4)}{(2n+3)(2n+5)} \end{align}$$

Putting $n=20$: $$\begin{align} \sum_{i=1}^{20}a_i &=\frac{20\cdot 24}{43\cdot 45}=\frac{32}{129}\qquad\blacksquare \end{align}$$

Answer 3

Alpha finds the sum to be $\frac {32}{129}$, giving an answer to your question of $1$. The way to do it by hand is telescoping the series.