I am not really sure what I was expecting when I typed in Sum[n/((n + 1) (n + 2) (n + 3)), {n, 1, x}] into WolframAlpha, but it strangely simplified it to: $$\sum_{n=1}^{x}\frac{n}{(n+1)(n+2)(n+3)}=\frac{x(x+1)}{4(x+2)(x+3)}$$ I feel that since the RHS is so simple, I must be overthinking it, but I can not figure out how WolframAlpha simplified this. Thanks in advance for insight.

• Have you tried using partial fractions on the summand? – JimmyK4542 Nov 30 '15 at 0:30
• @JimmyK4542 Yeah, I got$$\frac{-0.5}{n+1}+\frac{2}{n+2}+\frac{-1.5}{n+3}$$ but I wasn't sure where to go from there. – Rremmuoo Nov 30 '15 at 0:36

Your partial fraction has the clue.

$$2\sum_{n=1}^x\frac{n}{(n+1)(n+2)(n+3)}\\ =\sum_{n=1}^x\left(\frac{-1}{n+1}+\frac{4}{n+2}+\frac{-3}{n+3}\right)\\ =\sum_{n=2}^{x+1}\frac{-1}{n}+\sum_{n=3}^{x+2}\frac{4}{n}+\sum_{n=4}^{x+3}\frac{-3}{n}\\ =\left(-\frac{1}{2}-\frac{1}{3}+\sum_{n=4}^{x+1}\frac{-1}{n}\right)+\left(\frac{4}{3}+\frac{4}{x+2}+\sum_{n=4}^{x+1}\frac{4}{n}\right)+\left(-\frac{3}{x+2}-\frac{3}{x+3}+\sum_{n=4}^{x+1}\frac{-3}{n}\right)\\ =-\frac{1}{2}-\frac{1}{3}+\frac{4}{3}+\frac{4}{x+2}-\frac{3}{x+2}-\frac{3}{x+3}$$

• The 2 in the front threw me off at first, but now I get it was to make the -0.5 and -1.5 look better. Thanks! – Rremmuoo Nov 30 '15 at 1:06

Using partial fractions, you get that $\dfrac{n}{(n+1)(n+2)(n+3)} = \dfrac{-\tfrac{1}{2}}{n+1} + \dfrac{2}{n+2} + \dfrac{-\tfrac{3}{2}}{n+3}$.

Now, write out the terms being summed up:

$\left(\dfrac{-\tfrac{1}{2}}{2} + \dfrac{2}{3} + \dfrac{-\tfrac{3}{2}}{4}\right)+$ $\left(\dfrac{-\tfrac{1}{2}}{3} + \dfrac{2}{4} + \dfrac{-\tfrac{3}{2}}{5}\right)+$ $\left(\dfrac{-\tfrac{1}{2}}{4} + \dfrac{2}{5} + \dfrac{-\tfrac{3}{2}}{6}\right)+ \cdots$

$\left(\dfrac{-\tfrac{1}{2}}{x-1} + \dfrac{2}{x} + \dfrac{-\tfrac{3}{2}}{x+1}\right)+$ $\left(\dfrac{-\tfrac{1}{2}}{x} + \dfrac{2}{x+1} + \dfrac{-\tfrac{3}{2}}{x+2}\right)+$ $\left(\dfrac{-\tfrac{1}{2}}{x+1} + \dfrac{2}{x+2} + \dfrac{-\tfrac{3}{2}}{x+3}\right)$

We can regroup the terms as follows:

$\left(\dfrac{-\tfrac{1}{2}}{2}\right)$ $+\left(\dfrac{2}{3}+\dfrac{-\tfrac{1}{2}}{3}\right)$ $+\left(\dfrac{-\tfrac{3}{2}}{4}+\dfrac{2}{4}+\dfrac{-\tfrac{1}{2}}{4}\right)$ $+\left(\dfrac{-\tfrac{3}{2}}{5}+\dfrac{2}{5}+\dfrac{-\tfrac{1}{2}}{5}\right) + \cdots$

$+ \left(\dfrac{-\tfrac{3}{2}}{x+1}+\dfrac{2}{x+1}+\dfrac{-\tfrac{1}{2}}{x+1}\right)$ $+\left(\dfrac{-\tfrac{3}{2}}{x+2}+\dfrac{2}{x+2}\right)$ $+\left(\dfrac{-\tfrac{3}{2}}{x+3}\right)$

Each of the groups of terms, except the first two and the last two, sum to zero. So, we are left with:

$\left(\dfrac{-\tfrac{1}{2}}{2}\right)$ $+\left(\dfrac{2}{3}+\dfrac{-\tfrac{3}{2}}{3}\right)$ $+\left(\dfrac{-\tfrac{3}{2}}{x+2}+\dfrac{2}{x+2}\right)$ $+\left(\dfrac{-\tfrac{3}{2}}{x+3}\right)$ $= \dfrac{1}{4} + \dfrac{\tfrac{1}{2}}{x+2} + \dfrac{-\tfrac{3}{2}}{x+3}$

Combining those fractions will give you the result.

Note: This type of sum is known as a telescoping sum.

• Wow, if I actually took my time to just expand the series, I would've noticed that. I apologize for wasting your time, but now I know of telescoping sums thanks to you. – Rremmuoo Nov 30 '15 at 1:12