Question

\( 6 \quad T_{n}=\frac{n+2}{n(n+1)(n+3)} \) find \( S_{\infty} \)

Answer 1

Tn=         (n+3-1)/n(n+1)(n+3)
    =           1/n(n+1) -   1/n(n+1)(n+3)
    =           1/n(n+1) -   1/n(n+1)((n+2)+1)
    =           1/n(n+1) -   1/n(n+1)(n+2) - 1/n(n+1)(1)

    =           -   1/n(n+1)(n+2)
    =           -1/2[( n+2 -n)/n(n+1)(n+2)]
    =          -1/2[1/n(n+1) - 1/(n+1)(n+2)]
asked us about summation from 1 to infinity:

    =         -1/2[ 1/2 -  1/infinity]

    = -1/4