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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 $$\sum_{r=2}^\infty T_r$$

Michael Hardy
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Rohan Shinde
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    Did you start by considering $T_2$ followed by $T_3$? – abiessu Nov 15 '17 at 18:33
  • Sorry I have now edited the question – Rohan Shinde Nov 15 '17 at 18:36
  • I don't know immediately how to decompose that expression into partial fractions, but I would start just the way abiessu suggests, doing it with $T_2$ and $T_3$ and maybe $T_4$ or further. If I see a pattern, I might try to prove by mathematical induction on $r$ that the pattern continues. – Michael Hardy Nov 15 '17 at 18:39
  • Can you please elaborate your method I couldn't understand what you mean to say about induction – Rohan Shinde Nov 15 '17 at 18:44
  • So this one does telescope, it just takes some work to find the right decomposition... – abiessu Nov 16 '17 at 19:07

3 Answers3

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$$ T_{r}(x) = \frac{rx}{\displaystyle{\prod_{k=1}^{r}(1-kx)}} $$

The main idea is to decompose like this: $$\frac{rx}{\displaystyle{\prod_{k=1}^{r}(1-kx)}} =\frac{A}{\displaystyle{\prod_{i=1}^{r-1}(1-ix)}}+\frac{B}{\displaystyle{\prod_{j={2}}^{r}(1-jx)}}$$ $$=\frac{A}{(1-x)(1-2x)\cdots(1-(r-1)x)}+\frac{B}{(1-2x)(1-3x)\cdots(1-rx)} $$ $$ \frac{rx}{\displaystyle{\prod_{k=1}^{r}(1-kx)}} = \frac{A}{\displaystyle{\prod_{i=1}^{r-1}(1-ix)}}+\frac{B}{\displaystyle{\prod_{j=2}^{r}(1-jx)}} =\frac{A\cdot (1-rx)}{(1-rx)\displaystyle{\prod_{i=1}^{r-1}(1-ix)}}+\frac{B\cdot (1-x)}{\displaystyle{(1-x)\prod_{j=2}^{r}(1-jx)}} = \frac{(-B-rA)x+(A+B)}{\displaystyle{\prod_{k=1}^{r}(1-kx)}}$$

Then

$$A=\frac{r}{1-r}$$ $$B=\frac{-r}{1-r}$$

At least you can decompose as follows:

$$ T_{r}(x)= \frac{r}{1-r}\left(\frac{1}{\displaystyle{\prod_{j=1}^{r-1}(1-jx)}}-\frac{1}{\displaystyle{\prod_{j=2}^{r}(1-jx)}}\right) $$ $$T_{r}(x)=\left(\frac{\displaystyle{\frac{r}{1-r}}}{\displaystyle{\prod_{j=1}^{r-1}(1-jx)}}-\frac{\displaystyle{\frac{r}{1-r}}}{\displaystyle{\prod_{j=2}^{r}(1-jx)}}\right)$$

Hector Blandin
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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:

$$\begin{align}T_2&=\frac {2x}{(1-x)(1-2x)}=\frac A{1-2x}+\frac B{1-x}\\ 2x &= A(1-x)+B(1-2x)\to A=-B=2\\ T_2&=\frac 2{1-2x}-\frac 2{1-x}\\ T_3=\frac 32T_2\frac 1{1-3x}&=\frac 3{(1-2x)(1-3x)}-\frac 3{(1-x)(1-3x)}\\ &=\frac 9{1-3x}-\frac 6{1-2x}-\frac 9{2(1-3x)}+\frac 3{2(1-x)}\\ T_4=\frac 43T_3\frac 1{1-4x}&=\frac 6{(1-3x)(1-4x)}-\frac 8{(1-2x)(1-4x)}+\frac 2{(1-x)(1-4x)}\\ &=\dots \end{align}$$

abiessu
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This simplifies down pretty nicely with an algebra package. I think the other answers give you the right approach to attempt this by hand.

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for x<>0.

Kelly Lowder
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  • This is false for the partial sums up to $T_4$, and perhaps the algebra package is making a mistake similar to what is seen below, that $T_{r+1}$ does not equate to a product starting at $j=2$. Or perhaps the full sum really does converge nicely, but not in a telescoping manner... – abiessu Nov 15 '17 at 20:24
  • The full sum really does converge nicely. – Kelly Lowder Nov 15 '17 at 23:14
  • I think I can see it, that all the numerators go to $0$ except the $\frac A{1-x}$ term... – abiessu Nov 16 '17 at 00:02