Prove that the following sum converges and has the given value.
![\[ \sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)} = \frac{1}{4}. \]](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-3aa36ace5423321ad3f7e585a1071898_l3.png "Rendered by QuickLaTeX.com")
So, first we need to use partial fractions to decompose the terms in the sum,
![\[ \frac{n}{(n+1)(n+2)(n+3)} = \frac{A}{(n+1)(n+2)} + \frac{B}{(n+2)(n+3)}. \]](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-ee03e2d7096c76a36066f551e82ad66c_l3.png "Rendered by QuickLaTeX.com")
This gives us the equations,
![\[ A(n+2)(n+3) + B(n+1)(n+2) = n \quad \implies \quad A = -\frac{1}{2}, \ \ B = \frac{3}{2}. \]](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-2bc1255e3819736254c6217f8bc623a4_l3.png "Rendered by QuickLaTeX.com")
Therefore we have,
![\begin{align*} \sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)} &= \frac{1}{2} \left( \sum_{k=1}^{\infty} \left( \frac{-1}{(n+1)(n+2)} + \frac{3}{(n+2)(n+3)} \right) \\[9pt] &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \left( \frac{1}{(n+2)(n+3)} - \frac{1}{(n+1)(n+2)} \right) + \frac{2}{(n+2)(n+3)} \right) \\[9pt] &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{(n+2)(n+3)} - \frac{1}{(n+1)(n+2)} + \frac{A}{n+2} + \frac{B}{n+3} \right). \end{align*}](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-c62c80dbac4469d0cb7ba4f5950780fb_l3.png "Rendered by QuickLaTeX.com")
Using partial fraction decomposition again, we have 
and 
in the above equation, so we have
![\begin{align*} &= -\frac{1}{2} \sum_{k=1}^{\infty} \left( \left( \frac{1}{(n+1)(n+2)} - \frac{2}{n+2} \right) - \left( \frac{1}{(n+2)(n+3)} - \frac{2}{n+3} \right) \right)\\[9pt] &= -\frac{1}{2} \sum_{n=1}^{\infty} b_n - b_{n+1} \end{align*}](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-f9f1f163975fdc7998168b2356c6144f_l3.png "Rendered by QuickLaTeX.com")
where 
. Therefore, by the theorem on telescoping sums (Theorem 10.7 of Apostol) we have
![\begin{align*} \sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)} &= -\frac{1}{2} (b_1 - L) \\[9pt] &= -\frac{1}{2} \left( \frac{1}{6} - \frac{2}{3} - 0 \right) \\[9pt] &= \frac{1}{4}. \end{align*}](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-6e1b7207994fc69a2070e7159ce074f5_l3.png "Rendered by QuickLaTeX.com")
I always thought partial fraction decomposition required the denominator to be irreducible yet you have reducible quadratics; i.e. shouldn’t the numerators be in the AX+B format? Hopefully learning linear algebra down the road will clear this up…