<s>[INST] Sum of infinite sequence Let $$T_r=\frac{rx}{(1-x)(1-2x)(1-3x)\cdots(1-rx)}$$ 
 Can someone please tell me how to break this expression into partial fractions (because I am a bit weak at it) to find the following $$\sum_{r=2}^\infty T_r$$
 [/INST] This sum is telescopic with a different decomposition as follows:
$$\begin{align}
T_r&=\frac{rx}{\prod\limits_{k=1}^r1-kx}=\frac{1}{\prod\limits_{k=1}^{r}1-kx}-\frac{1-rx}{\prod\limits_{k=1}^{r}1-kx}=\frac{1}{\prod\limits_{k=1}^{r}1-kx}-\frac{1}{\prod\limits_{k=1}^{r-1}1-kx}
\end{align}$$
It should be fairly clear that $-\frac 1{1-x}$ stays and all other terms telescope to $0$.  It is also interesting to note that if we begin the sum at $r=1$, we get the result $-1={x-1\over1-x}$.  For an approach on how to decompose all the terms directly, read on:
$$\begin{align}T_2&=\frac {2x}{(1-x)(1-2x)}=\frac A{1-2x}+\frac B{1-x}\\
2x &= A(1-x)+B(1-2x)\to A=-B=2\\
T_2&=\frac 2{1-2x}-\frac 2{1-x}\\
T_3=\frac 32T_2\frac 1{1-3x}&=\frac 3{(1-2x)(1-3x)}-\frac 3{(1-x)(1-3x)}\\
&=\frac 9{1-3x}-\frac 6{1-2x}-\frac 9{2(1-3x)}+\frac 3{2(1-x)}\\
T_4=\frac 43T_3\frac 1{1-4x}&=\frac 6{(1-3x)(1-4x)}-\frac 8{(1-2x)(1-4x)}+\frac 2{(1-x)(1-4x)}\\
&=\dots
\end{align}$$
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