An algebra problem by Pi Han Goh

Let $\displaystyle T_{r} = \frac{1}{r(r+1)(r+2)(r+3)(r+4)(r+5)}$ .
Therefore, $\displaystyle A = \sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)(n+3)(n+4)(n+5)} = \sum_{n=1}^\infty T_{n}$
Let us define a new sequence of variables, $a_{r}$ .
$\displaystyle a_{r} = \frac{1}{r(r+1)(r+2)(r+3)(r+4)}$
$\displaystyle \Rightarrow a_{r+1} = \frac{1}{(r+1)(r+2)(r+3)(r+4)(r+5)}$
Hence,
$\displaystyle a_{r} - a_{r+1} = \frac{1}{r(r+1)(r+2)(r+3)(r+4)} - \frac{1}{(r+1)(r+2)(r+3)(r+4)(r+5)}$

$\displaystyle \Rightarrow a_{r} - a_{r+1} = \frac{r+5-r}{r(r+1)(r+2)(r+3)(r+4)(r+5)}$

$\displaystyle \Rightarrow a_{r} - a_{r+1} = 5 \times \frac{1}{r(r+1)(r+2)(r+3)(r+4)(r+5)} = 5T_{r}$

Therefore, we can write,
$\displaystyle A = \sum_{n=1}^\infty T_{n} = \frac{1}{5} \times \sum_{n=1}^\infty (a_{n} - a_{n+1})$ , which being an easy telescopic series evaluates to $\displaystyle \frac{a_{1}}{5} = \frac{1}{600}$

Therefore, $\displaystyle \boxed{\frac{1}{A} = 600}$

$\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)(n+2)(n+3)(n+4)(n+5)}$

= $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{5} \bigg( \dfrac{(n+5) - n}{n(n+1)(n+2)(n+3)(n+4)(n+5)} \bigg)$

= $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{5}\bigg(\dfrac{1}{n(n+1)(n+2)(n+3)(n+4)} - \dfrac{1}{(n+1)(n+2)(n+3)(n+4)(n+5)} \bigg)$

= $\dfrac{1}{5 \times 5!} - 0 = \dfrac{1}{600}$ (using astronomy, i.e. telescoping series :P)

$\color{#20A900}{ A = \displaystyle \sum_{n=1}^\infty \frac 1{n(n+1)(n+2)(n+3)(n+4)(n+5)} < ~\sum_{n=1}^\infty \frac 1{n^2}, ~both~are~convergent.\\ A = \displaystyle \sum_{n=1}^\infty\Big \{ \frac B n+\frac D{n+1}+\frac F{n+2}+\frac G{n+3}+\frac E{n+4}+\frac C {n+5} \Big \} .\\ Using~ cover~ up ~rule:-\\ B=+\frac 1 {120},~~~~D= - \frac 5 {120},~~~F=+\frac {10} {120},\\ C= - \frac 1 {120},~~~E=+ \frac 5 {120},~~~G= - \frac {10}{120}\\ \therefore~A = \displaystyle ~\frac 1 {120}~ \sum_{n=1}^\infty~\Bigg \{ \bigg(\frac 1 n-\frac 1{n+5}\bigg)-\bigg(\frac 5{n+1}-\frac 5{n+4} \bigg)+\bigg(\frac {10}{n+2}-\frac {10} {n+3}\bigg) \Bigg \} , \\ \therefore~A = \displaystyle ~\frac 1 {120}~ ~\Big \{ \sum_{n=1}^5 \frac 1 n - \sum_{n=1}^3\frac 5{n+1} + \sum_{n=1}^1\frac {10}{n+2} \Big \} \\ \displaystyle ~~+\sum_{n=1}^5 \frac 1 n~\quad \quad \quad=~~~~~~~~~\dfrac 1 1~\quad~~~~~\dfrac 1 2~~\quad~~~~\dfrac 1 3~~~\quad~~~\dfrac 1 4~~~\quad~~~\dfrac 1 5 \\ \displaystyle ~~-\sum_{n=1}^3\frac 5{n+1}~\quad~~~=~\quad~\quad~\quad~-\dfrac 52~\quad~-\dfrac 53~\quad~-\dfrac 5 4\\ \displaystyle ~+ \sum_{n=1}^1\frac {10}{n+2}\quad~\quad~=~~~\quad~~\quad\quad~~~~~~~~~~~~~~~~\dfrac {10} 3 \\ Finally~120*A=\displaystyle ~~+\sum_{n=1}^5 \frac 1 n - \sum_{n=1}^3\frac 5{n+1} + \sum_{n=1}^1\frac {10}{n+2}=\frac 1 5. \\ \therefore~~\dfrac 1 A=} \Huge \color{#D61F06}{600}. \\$

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$\color{#EC7300}{ *~~Using~ cover~ up ~rule:-\\ \displaystyle B_{n=~0} =\frac 1 {1*2*3*4*5}=~~~~~~~~~~~~+\frac {~1} {120}\\ \displaystyle D_{n=-1} =\frac 1 {-1*1*2*3*4}=~~~~~~~~ - \frac{~ 5} {120}\\ \displaystyle F_{n=-2} =\frac 1 {-2*-1*1*2*3}=~~~~ ~+ \frac {10} {120}\\ \displaystyle G_{n=-3} =\frac 1 {-3*-2*-1*1*2}= ~~- \frac {10} {120}\\ \displaystyle E_{n=-4} =\frac 1 {-4*-3*-2*-1*1}=~+\frac{~ 5} {120}\\ \displaystyle C_{n=-5} =\frac 1 {-5*-4*-3*-2*-1}= - \frac 1 {120} }\\ \color{#3D99F6}{* * \displaystyle \sum_{i=m}^\infty f(i) = \sum_{i=m-k}^\infty f(i+k)~~~~ \\ \therefore~\displaystyle \sum_{n=1}^\infty \Big ( \frac1 n-\frac 1{n+5} \Big)= \Big ( \sum_{n=1}^5 \frac1 n+\sum_{n=6}^\infty \frac1 n \Big ) - \sum_{n=1}^\infty\frac 1{n+5} \\ \displaystyle = \sum_{n=1}^5 \frac1 n + \Big ( \sum_{n=1}^\infty\frac 1{n+5} - \sum_{n=1}^\infty\frac 1{n+5} \Big) ~~~~~and~the~original~~ series ~~is ~convergent,\\ \displaystyle = \sum_{n=1}^5 \frac1 n .\\ Similarly~~\displaystyle~ \sum_{n=1}^\infty \Big \{\frac 5{n+1}-\frac 5{n+4} \Big \}= \sum_{n=1}^3 \Big (\frac 5{n+1} \Big ) ~\\ and ~ \displaystyle~ \sum_{n=1}^\infty \Big \{\frac {10}{n+2}-\frac{10} {n+3} \Big \}= \sum_{n=1}^1 \Big (\frac {10}{n+3} \Big ) }$