Home » Blog » Prove that the sum from 1 to ∞ of n / ((n+1)(n+2)(n+3)) = 1/4

Prove that the sum from 1 to ∞ of n / ((n+1)(n+2)(n+3)) = 1/4

by RoRi

Prove that the following sum converges and has the given value.

    \[ \sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)} = \frac{1}{4}. \]

So, first we need to use partial fractions to decompose the terms in the sum,

    \[ \frac{n}{(n+1)(n+2)(n+3)} = \frac{A}{(n+1)(n+2)} + \frac{B}{(n+2)(n+3)}. \]

This gives us the equations,

    \[ A(n+2)(n+3) + B(n+1)(n+2) = n \quad \implies \quad A = -\frac{1}{2}, \ \ B = \frac{3}{2}. \]

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*}

Using partial fraction decomposition again, we have A = -2 and B = 2 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*}

where b_n = \frac{1}{(n+1)(n+2)} - \frac{2}{n+2}. 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*}

7 comments

  1. tom says:

    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…

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