This question is similarly found here, and I have been trying to break it down using partial fraction however I cannot seem to get the final result as in the thread linked. I would very much appreciate some advice on my solution!
$\sum_{n=1}^{\infty} \frac{n}{(n+1)(n+2)(n+3)} = \frac{1}{4}$
I have tried splitting this into Partial fractions:
$$n = \frac{A}{(n+2)(n+3)}+\frac{B}{(n+1)(n+3)}+\frac{C}{(n+1)(n+2)}$$
when n = 1 I find that A = $\frac{1}{12}$; When n = 2 then B = $\frac{2}{15}$; and when n = 3 then C = $\frac{3}{17}$
This then gets me:
$$\frac{(1/12)}{(n+1)}+\frac{(2/15)}{(n+2)}+\frac{(3/17)}{(n+3)}$$
However, I cannot seem to derive the solution $\frac{1}{4}$ from this.
I think you may have computed the partial fractions incorrectly. If we write $$\frac{n}{(n+1)(n+2)(n+3)} = \frac A{n+1} + \frac B{n+2} + \frac C{n+3}$$ then this rearranges as $$n = A(n+2)(n+3) + B(n+1)(n+3) + C(n+1)(n+2)$$ and we need to plug in $n = -1, -2, -3$. Setting $n = -1$ gives us $A = -\frac12$, setting $n = -2$ gives us $B=2$, and setting $n = -3$ gives us $C = -\frac32$, so
\begin{align} \frac{n}{(n+1)(n+2)(n+3)} & = -\frac{1/2}{n+1} + \frac{2}{n+2} - \frac{3/2}{n+3}\\ & = -\frac{1/2}{n+1} + \frac{1/2}{n+2} + \frac{3/2}{n+2} - \frac{3/2}{n+3} \\ & = \dfrac12\left(\dfrac{1}{n+2}-\dfrac{1}{n+1}\right)+\dfrac32\left(\dfrac{1}{n+2}-\dfrac{1}{n+3}\right) \end{align}
and now the telescoping is more clear (this is the same form as in E.H.E.'s answer in the post you linked).