Question

Sum the following up to infinity?

`1/(2*5)+1/(3*6)+1/(4*7)+1/(5*8)+1/(6*9)+.....`

Answer 1

`13/36`

Explanation:

convert it into this:-

`1/3{[1/2-1/5]+[1/3-1/6]+[1/4-1/7]+......}`

then, divide the sum into to parts- the positive constants and the negative constants.

`=1/3{[1/2+1/3+1/4+1/5+......]-[1/5+1/6+1/7+......]}`

cancel all the constants of opposite signs,i.e., from `1/5` to infinity.

`=1/3{1/2+1/3+1/4}`

take LCM on the inside

`=1/3{[6+4+3]/12}`

`=13/36`

hope this helps,
Shivang M.

Answer 2

`sum_(n=1)^oo 1/((n+1)(n+4)) = 13/36`

Explanation:

`1/(2*5)+1/(3*6)+1/(4*7)+...= sum_(n=1)^oo 1/((n+1)(n+4))`

Note that:

`1/((n+1)(n+4)) = 1/3(1/(n+1)-1/(n+4))`

So:

`sum_(n=1)^N 1/((n+1)(n+4))`

`=1/3 sum_(n=1)^N (1/(n+1)-1/(n+4))`

`=1/3 (sum_(n=1)^N 1/(n+1)- sum_(n=1)^N 1/(n+4))`

`=1/3 (sum_(n=1)^N 1/(n+1)- sum_(n=4)^(N+3) 1/(n+1))`

`=1/3 (1/2+1/3+1/4+color(red)(cancel(color(black)(sum_(n=4)^N 1/(n+1))))- color(red)(cancel(color(black)(sum_(n=4)^N 1/(n+1))))-1/(N+2)-1/(N+3)-1/(N+4))`

`=1/3 (1/2+1/3+1/4-1/(N+2)-1/(N+3)-1/(N+4))`

Then:

`sum_(n=1)^oo 1/((n+1)(n+4)) = lim_(N->oo) sum_(n=1)^N 1/((n+1)(n+4))`

`color(white)(sum_(n=1)^oo 1/((n+1)(n+4))) = lim_(N->oo) 1/3 (1/2+1/3+1/4-1/(N+2)-1/(N+3)-1/(N+4))`

`color(white)(sum_(n=1)^oo 1/((n+1)(n+4))) = 1/3 (1/2+1/3+1/4)`

`color(white)(sum_(n=1)^oo 1/((n+1)(n+4))) = 1/3 ((6+4+3)/12)`

`color(white)(sum_(n=1)^oo 1/((n+1)(n+4))) = 13/36`