How do I do the following series:?

Question

Examine the series: Sum of 1/n(n+1)(n+2).

How does this converge to 1/4? Thanks.

1 / [n(n+1)(n+2)]

I know in this form it's divergent as the harmonic series, so how does it converge to 1/4.

Answer 1

Using method of partial fractions:

1/n(n+1)(n+2) = 0.5*(1/n) - 1/(n+1) + 0.5*(1/(n+2))

The sum of this term from 1 to m is:

{ 0.5*(1/1) - 1/(2) + 0.5*(1/(3)) } + { 0.5*(1/2) - 1/(3) + 0.5*(1/(4)) } ... + { 0.5*(1/m) - 1/(m+1) + 0.5*(1/(m+2))

=

0.5*(1/1) + (0.5-1)*(1/2) + (0.5-1+0.5)*(1/3) + (0.5-1+0.5)*(1/4) ... + (0.5-1+0.5)*(1/m) + (-1+0.5)*(1/(m+1)) + (0.5)*(1/(m+2))

=

0.5 - 0.25 - 0.5*(1/(m+1)) + 0.5*(1/(m+2))

as m approaches infinity 1/(m+1) and 1/(m+2) approaches zero.

The sum approaches 0.5-0.25=1/4

Answer 2

What series? All you have is an unclear expression.
Is it (n+1)(n+2)/n? or 1/[n(n+1)(n+2)]?

Restate the problem so we can help you.