माना कि श्रेणी का n वां पद `t_(n)" और "n` पदों का योगफल `S_(n)` है ।
`:.t_(n)=(1)/((1,2,3........."का n वां पद ")xx(2,3,4,.........."का n वां पद ")xx(3,4,5,......."का n वां पद"))`
`=(1)/(n(n+1)(n+2))`
माना कि `t_(n)=(A)/(n)+(B)/(n+1)+(C)/(n+2)" ".........(1)`
तो `A=(1)/((n+1)(n+2))` का मान जब `n=0=(1)/((0+1)(0+2))=(1)/(2)`
`B=(1)/(n(n+2))` का मान जब n+1=0 अर्थात जब n=-1
`=(1)/(-1(-1+2))=-1`
`C=(1)/(n(n+1))` का मान जब n+2=0 अर्थात जब n=-2
`=(1)/(-2(-2+1))=(1)/(2)`
(1) से, `t_(n)=(1)/(2n)-(1)/(n+1)+(1)/(2(n+2))`
n=1, 2, 3,........... रखने पर
`t_(1)=(1)/(2.1)-(1)/(2)+(1)/(2.3)`
`t_(2)=(1)/(2.2)-(1)/(3)+(1)/(2.4)`
`t_(3)=(1)/(2.3)-(1)/(4)+(1)/(2.5)`
`........................`
`..................................`
`t_(n-2)=(1)/(2(n-2))-(1)/(n-1)+(1)/(2n)`
`t_(n-1)=(1)/(2(n-1))-(1)/(n)+(1)/(2(n+1))`
`t_(n)=(1)/(2n)-(1)/(n+1)+(1)/(2(n+2))`
जोड़ने पर, `S_(n)=(1)/(2.1)-(1)/(2)+(1)/(2.2)+(1)/(2(n+1))-(1)/(n+1)+(1)/(2(n+1))`
`=(1)/(4)-(1)/(2(n+1))+(1)/(2(n+2))`
`:.S_(oo)` (sum to infinity) `=underset(ntooo)(LtS_(n))`
`=underset(ntooo)(Lt)[(1)/(4)-(1)/(2(n+1))+(1)/(2(n+2))]=(1)/(4)`
