Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$

Question

Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$

My method:

I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in $A,B,C,D$! And solving them takes much time. I split into partial fractions hoping that some of them will cancel out and the sum might telescope.

I hope theres a simpler method to this.

Edit

As a side-question, can we extend this to $k$ factors in denominator, something like

$$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)...(r+(k-1)d)}$$

Using N.S answer, we can extend it easily:

$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$

Answer 1

And solving them takes much time.

There is a trick so solving partial fractions quickly.

$$\frac{1}{(r)(r + d)(r + 2d)(r + 3d)} = \frac{A}{r} + \frac{B}{r + d} + \frac{C}{r + 2d} + \frac{D}{r + 3d}$$

$$1 = A(r + d)(r + 2d)(r + 3d) + B(r)(r + 2d)(r + 3d) + C(r)(r + d)(r + 3d) + D(r)(r + d)(r + 2d)$$

The above is true for all $r$. So plug in the roots of the original equation. $r = 0$ gives:

$$1 = A(d)(2d)(3)$$ $$A = \frac{1}{6d^3}$$

Then $r = -d$ :

$$1 = B(-d)(d)(2d)$$ $$B = \frac{-1}{2d^3}$$

Then $r = -2d$:

$$1 = C(-2d)(-d)(d)$$ $$C = \frac{1}{2d^3}$$

Last $r = -3d$:

$$1 = D(-3d)(-2d)(-d)$$ $$D = \frac{-1}{6d^3}$$

So you get:

$$S = \frac{1}{6d^3}\left( \frac{1}{r} - \frac{3}{r + d} + \frac{3}{r + 2d} - \frac{1}{r + 3d}\right)$$

After you do it a few times and get the pattern, you can almost do it in your head. I suggest looking at it as 2 telescoping series:

$$\text{Sum} = \left(\frac{1}{6d^3} \sum_{r = 1}^n \frac{1}{r} - \frac{1}{r + 3d}\right) + \left(\frac{-1}{2d^3} \sum_{r = 1}^n \frac{1}{r + d} - \frac{1}{r + 2d}\right)$$

Answer 2

HINT: $$\frac{1}{r(r+d)(r+2d)(r+3d)} = \frac{1}{2d^2}\left(\frac{1}{r(r+3d)}-\frac{1}{(r+d)(r+2d)}\right) = \frac{1}{6d^3}\left(\frac{1}{r}-\frac{1}{r+3d}\right)-\frac{1}{2d^3}\left(\frac{1}{r+d}-\frac{1}{r+2d}\right).$$

Answer 3

Hint

$$\frac{1}{(r)(r+d)(r+2d)(r+3d)}=\frac{1}{3d}\left(\frac{1}{(r)(r+d)(r+2d)}-\frac{1}{(r+d)(r+2d)(r+3d)}\right)$$

Answer 4

When computing a partial fraction decomposition one can always invoke residues. In our case $$ f(x)=\frac{1}{x(x+d)(x+2d)(x+3d)} = \frac{A_0}{x}+\frac{A_1}{x+d}+\frac{A_2}{x+2d}+\frac{A_3}{x+3d} $$ and $$ A_k = \text{Res}\left(f(x),x=-kd\right) = \lim_{x\to -kd}(x+kd)\,f(x)=\frac{1}{\prod_{\substack{0\leq j\leq 3 \\ j\neq k}}(-kd+jd)}$$ so $(A_0,A_1,A_2,A_3)$ is proportional to $\left(+\binom{3}{0},-\binom{3}{1},+\binom{3}{2},-\binom{3}{3}\right)$.
This approach has a great generality and it clearly shows that the coefficients appearing in the partial fraction decomposition of $\frac{1}{x(x+d)\cdots(x+Md)}$ are proportional to the binomial coefficients $\binom{M}{k}$ took with alternating signs.

Answer 5

(Modified answer)

Define $r^{\overline{m}(d)}$ as the $d$-tuple rising factorial, i.e. $r^{\overline{m}(d)}=r(r+d)(r+2d)\cdots(r+(m-1)d) $ and $r^{-\overline{m}(d)}=\frac 1{r^{\overline{m}(d)}}$ , e.g. $r^{\overline{4}(3)}=r(r+3)(r+6)(r+9)$.

$$\begin{align} S(m,d)&=\sum_{r=1}^n\frac 1{r(r+d)(r+2d)\cdots (r+(m-1)d)}\\ &=\sum_{r=1}^n r^{-\overline{m}(d)}\\ &=\frac 1{(m-1)d}\sum_{r=1}^n r^{-\overline{m-1}(d)}\ -\ (r+d)^{-\overline{m-1}(d)}\\ &=\boxed{\frac 1{(m-1)d}\sum_{r=1}^d r^{-\overline{m-1}(d)}\ - (n+r)^{-\overline{m-1}(d)} } \end{align}$$

For example, if $d=3, m=4, n=100$, $$\small\begin{align} &S(3,4)\\ &=\sum_{r=1}^{100}\frac 1{r(r+3)(r+6)(r+9)}\\ &\color{lightgrey}{=\small\frac 1{1\cdot 4\cdot7\cdot 10} +\frac 1{2\cdot5\cdot8\cdot 11}+\frac 1{3\cdot6\cdot9\cdot 12} +\cdots+ \frac 1{100\cdot103\cdot106\cdot109} }\\ &=\frac 19 \sum_{r=1}^{100}\frac 1{r(r+3)(r+6)}-\frac 1{(r+3)(r+6)(r+9)}\\ &=\frac 19\left[ \sum_{r=1}^{100}\frac 1{r(r+3)(r+6)}-\sum_{r=4}^{103}\frac 1{r(r+3)(r+6)}\right]\\ &=\frac 19\left[ \sum_{r=1}^3 \frac1{r(r+3)(r+6)}-\sum_{r=101}^{103}\frac 1{r(r+3)(r+6)(r+9)}\right]\\ &=\small\frac 19 \left[\frac 1{1\cdot 4\cdot7\cdot} +\frac 1{2\cdot5\cdot8}+\frac 1{3\cdot6\cdot9} -\frac 1{101\cdot 104\cdot 107} -\frac 1{102\cdot 105\cdot 108} -\frac 1{103\cdot106\cdot 109} \right]\end{align}$$