Sum the following up to infinity?

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

2 Answers
Jul 28, 2018

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.

Jul 28, 2018

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