An algebra problem by Christian Daang

$\displaystyle \sum_{n=4}^\infty \frac{1}{(n-2)(n-1)n (n+1)(n+2)}$
$= \frac{1}{4} \displaystyle \sum_{n=4}^\infty \frac{(n+2)-(n-2)}{(n-2)(n-1) n (n+1) (n+2)}$
$= \frac {1}{4} \displaystyle \sum_{n=4}^\infty\left( \frac {1}{(n-2)(n-1) n (n+1)} - \frac {1}{(n-1)n (n+1)(n+2)}\right)$

On observing we can see that on expanding the sum; everything cancels out except $\frac {1}{2*3*4*5}$
So, the sum is $\frac {1}{4} * \frac {1}{2*3*4*5} = \frac {1}{480}$
And our answer is $1+480=\boxed {481}$

Similar to Christian Daang 's approach.

$\begin{aligned} S & = \sum_{n=4}^\infty \frac 1{(n-2)(n-1)n(n+1)(n+2)} \quad \quad \quad \quad \small \color{#3D99F6} \text{By partial fractions -- see below.} \\ & = \sum_{n=4} ^\infty \left( \frac 1{24(n-2)} - \frac 1{6(n-1)} + \frac 1{4n} - \frac 1{6(n+1)} + \frac 1{24(n-2)} \right) \\ & = \frac 1{24} \sum_{n=4} ^\infty \left( \frac 1{n-2} - \frac 4{n-1} + \frac 6{n} - \frac 4{n+1} + \frac 1{n-2} \right) \\ & = \frac 1{24} \sum_{n=4} ^\infty \left({\color{#3D99F6}\frac 1{n-2} - \frac 1{n-1}}{\color{#D61F06} - \frac 3{n-1} + \frac 3{n}}{\color{#3D99F6}+ \frac 3{n} - \frac 3{n+1}}{\color{#D61F06} - \frac 1{n+1} + \frac 1{n-2}} \right) \\ & = \frac 1{24} \left({\color{#3D99F6}\frac 12}{\color{#D61F06} - \frac 33}{\color{#3D99F6}+ \frac 34}{\color{#D61F06} - \frac 15} \right) \\ & = \frac 1{24} \left(\frac {10-20+15-4}{20} \right) \\ & = \frac 1{480} \end{aligned}$

$\implies a + b = 1 + 480 = \boxed{481}$

---

Partial fractions:

Let $\dfrac 1{(n-2)(n-1)n(n+1)(n+2)} = \dfrac A{n-2} + \dfrac B{n-1} + \dfrac Cn + \dfrac D{n+1} + \dfrac E{n+2}$ .

Multiplying both sides with $(n-2)(n-1)n(n+1)(n+2)$ , we get:

$\small 1 = A(n-1)n(n+1)(n+2) + B{\color{#D61F06}(n-2)}n(n+1)(n+2) + C{\color{#D61F06}(n-2)}(n-1)(n+1)(n+2) + D{\color{#D61F06}(n-2)}(n-1)n(n+2) + E{\color{#D61F06}(n-2)}(n-1)n(n+1)$

If we put ${\color{#D61F06}n=2}$ , then we have $1=A(2-1)2(2+1)(2+2) + B{\color{#D61F06}(0)}+C{\color{#D61F06}(0)}+D{\color{#D61F06}(0)}+E{\color{#D61F06}(0)}$ .

$\small \begin{aligned} \implies \frac A{\color{#D61F06} \cancel{n-2}} & = \frac 1{({\color{#D61F06} \cancel{n-2}})(n-1)n(n+1)(n+2)} \bigg|_{\color{#D61F06}n=2} = \frac 1{(2-1)(2)(2+1)(2+2)} = \frac 1{(1)(2)(3)(4)} = \frac 1{24} \\ \frac B{\color{#D61F06} \cancel{n-1}} & = \frac 1{(n-2)({\color{#D61F06} \cancel{n-1}})n(n+1)(n+2)} \bigg|_{\color{#D61F06}n=1} = \frac 1{(1-2)(1)(1+1)(1+2)} = \frac 1{(-1)(1)(2)(3)} = - \frac 1 6 \\ \frac C{\color{#D61F06} \cancel n} & = \frac 1{(n-2)(n-1){\color{#D61F06} \cancel{n}}(n+1)(n+2)} \bigg|_{\color{#D61F06}n=0} = \frac 1{(0-2)(0-1)(0+1)(0+2)} = \frac 1{(-2)(-1)(1)(2)} = \frac 14 \\ \frac D{\color{#D61F06} \cancel{n+1}} & = \frac 1{(n-2)(n-1)n{(\color{#D61F06} \cancel{n+1}}) (n+2)} \bigg|_{\color{#D61F06}n=-1} = \frac 1{(-1-2)(-1-1)(-1)(-1+2)} = \frac 1{(-3)(-2)(-1)(1)} = - \frac 16 \\ \frac E{\color{#D61F06} \cancel {n+2}} & = \frac 1{(n-2)(n-1)n(n+1){(\color{#D61F06} \cancel{n+2}})} \bigg|_{\color{#D61F06}n=-2} = \frac 1{(-2-2)(-2-1)(-2)(-2+1)} = \frac 1{(-4)(-3)(-2)(-1)} = \frac 1{24} \end{aligned}$

Applying Partial Fractions, we can simplify it to:

$\begin{aligned} \sum_{n = 4}^\infty \left( \cfrac{1}{(n-2)(n-1)(n)(n+1)(n+2)} \right) & = \sum_{n = 4}^\infty \left( \cfrac{1}{(n-2)(24)} - \cfrac{1}{(n-1)(6)} + \cfrac{1}{(n)(4)} - \cfrac{1}{(n+1)(6)} + \cfrac{1}{(n+2)(24)} \right) \\ & = \cfrac{1}{24} \cdot \sum_{n = 4}^\infty \left( \cfrac{1}{(n-2)} - \cfrac{4}{(n-1)} + \cfrac{6}{(n)} - \cfrac{4}{(n+1)} + \cfrac{1}{(n+2)} \right ) \\ & = \cfrac{1}{24} \cdot \begin{pmatrix} \cfrac{1}{2} & - \cfrac{4}{3} & + \cfrac{6}{4} & - \cfrac{4}{5} & \color{#3D99F6} \cancel {+ \cfrac{1}{6}} \\ \cfrac{1}{3} & - \cfrac{4}{4} & + \cfrac{6}{5} & \color{#3D99F6} \cancel {- \cfrac{4}{6}} & \color{#3D99F6} \cancel {+ \cfrac{1}{7}} \\ \cfrac{1}{4} & - \cfrac{4}{5} & \color{#3D99F6} \cancel {+\cfrac{6}{6}} & \color{#3D99F6} \cancel {- \cfrac{4}{7}} & \color{#3D99F6} \cancel {+ \cfrac{1}{8}} \\ \cfrac{1}{5} & \color{#3D99F6} \cancel {- \cfrac{4}{6}} & \color{#3D99F6} \cancel {+ \cfrac{6}{7}} & \color{#3D99F6} \cancel {- \cfrac{4}{8}} & \color{#3D99F6} \cancel {+ \cfrac{1}{9}} \\ \color{#3D99F6} \cancel {\cfrac{1}{6}} & \color{#3D99F6} \cancel {- \cfrac{4}{7}} & \color{#3D99F6} \cancel {+ \cfrac{6}{8}} & \color{#3D99F6} \cancel {- \cfrac{4}{9}} & \color{#3D99F6} \cancel {+ \cfrac{1}{10}} \\ \color{#3D99F6} \cancel {\cfrac{1}{7}} & \color{#3D99F6} \cancel {- \cfrac{4}{8}} & \color{#3D99F6} \cancel {+ \cfrac{6}{9}} & \color{#3D99F6} \cancel {- \cfrac{4}{10}} & \color{#3D99F6} \cancel {+ \cfrac{1}{11}} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \color{#3D99F6} \cancel {\cfrac{1}{(n-2)}} & - \cfrac{4}{(n-1)} & + \cfrac{6}{(n)} & - \cfrac{4}{(n+1)} & + \cfrac{1}{(n+2)} \end{pmatrix} \end{aligned}$

As you notice, as the pattern will continues, the numbers colored in blue will be cancelled out and the sum will be: $\cfrac{1}{24} \cdot \left( \cfrac{1}{2} - \cfrac{4}{3} + \cfrac{1}{3} + \cfrac{6}{4} - \cfrac{4}{4} + \cfrac{1}{4} - \cfrac{4}{5} + \cfrac{6}{5} - \cfrac{4}{5} + \cfrac{1}{5} \right) - \cfrac{4}{(n-1)} + \cfrac{6}{(n)} - \cfrac{4}{(n+1)} + \cfrac{1}{(n+2)}$

which will converge to:

$\cfrac{1}{24} \cdot \left( \cfrac{1}{2} - \cfrac{4}{3} + \cfrac{1}{3} + \cfrac{6}{4} - \cfrac{4}{4} + \cfrac{1}{4} - \cfrac{4}{5} + \cfrac{6}{5} - \cfrac{4}{5} + \cfrac{1}{5} \right) = \cfrac{1}{480} \\ \implies \boxed{a + b = 481}$

$\sum _{ n=4 }^{ \infty }{ \frac { 1 }{ (n-2)(n-1)n(n+1)(n+2) } } =\quad \frac { 1 }{ 4 } \sum { \frac { 1 }{ (n-2)(n-1)(n)(n+1) } } -\frac { 1 }{ (n-1)n(n+1)(n+2) } =\frac { 1 }{ 4.2.3.4.5 } =\boxed { \frac { 1 }{ 480 } } \\ \{ \because \sum _{ k=a }^{ n }{ f\left( k \right) } -f\left( k+1 \right) \quad =\quad f\left( a \right) -f\left( n+1 \right) \}$