I'm having trouble solving this telescopic sum: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$$ I've split it into three fractions, trying to find a relationship that cancels out intermediate terms, but I'm stuck. Can anyone point me in the right direction? I basically split the fraction like this: $$\frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} = \frac{1}{n(n+1)(n+2)}$$
$$\frac{A(n+1) + Bn}{n(n+1)} + \frac{C}{n+2} = \frac{1}{n(n+1)(n+2)}$$
$$\frac{n(A+B) + A}{n(n+1)} + \frac{C}{n+2} = \frac{1}{n(n+1)(n+2)}$$
$$\frac{A(n+1) + Bn}{n(n+1)} + \frac{C}{n+2} = \frac{1}{n(n+1)(n+2)}$$
$$\frac{n(A+B) + A}{n(n+1)} + \frac{C}{n+2} = \frac{1}{n(n+1)(n+2)}$$
We have:
$\begin{pmatrix} \dot{x}_1 \\ \dot{x}_2\end{pmatrix} = \begin{pmatrix}3 & 1 \\ 4 & 3\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}+ \begin{pmatrix}0\\u \end{pmatrix}$
So we get eigenvalues from $(3-\lambda)^2-4=\lambda^2-6\lambda +5=(\lambda-1)(\lambda-5),\lambda=1,5$
Finding Eig...
