+9519 The sum, written in sigma notation, is \(\newcommand{\dsum}{\displaystyle\sum} \dsum_{k = 1}^{\infty} \dfrac2{k(k + 1)(k + 2)}\).
We can perform partial fractions to get a telescoping sum.
Note that \(\newcommand{\dsum}{\displaystyle\sum} \dsum_{k = 1}^\infty (f(k) - f(k + 1)) = f(1) - \lim_{k\to\infty} f(k)\) for any function f.
Let \(\dfrac2{k(k + 1)(k + 2)} = \dfrac A{k} + \dfrac B{k + 1} + \dfrac C{k + 2}\) for some constants A, B, C.
Multiplying both sides by k(k + 1)(k + 2) gives \(2 = A(k + 1)(k + 2) + Bk(k + 2) + Ck(k + 1)\).
Substituting k = 0, k = -1, and k = -2 respectively gives \(\begin{cases}2A = 2\\-B = 2\\2C = 2\end{cases}\).
Then A = C = 1, B = -2.
We have:
\(\newcommand{\dsum}{\displaystyle\sum} \begin{array}{rcl} \dsum_{k = 1}^{\infty} \dfrac2{k(k + 1)(k + 2)} &=& \dsum_{k = 1}^\infty \left(\dfrac1k + \dfrac1{k + 2} - \dfrac2{k + 1}\right)\\ &=& \dsum_{k = 1}^\infty \left(\dfrac1k - \dfrac1{k + 1}\right) - \dsum_{k = 1} ^\infty\left( \dfrac1{k + 1} - \dfrac1{k + 2}\right) \end{array}\)
Can you continue with my hints here?
