I’m not sure where I got this question from. It asks for the value of the infinite sum:

$\frac{3}{1\times 2 \times 3} + \frac{5}{2\times 3 \times 4} + \frac{7}{3\times4\times5} + \dots$

Have a go if you’d like; spoilers are below the line.

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The first thing I would do is work out the general term. The numerators are going up by two each time, so I reckon it’s $\frac{2k+1}{k(k+1)(k+2)}$, and we want

$S= \sum_{k=1}^{\infty} \frac{2k+1}{k(k+1){k+2}}$.

How do we get that? Partial fractions look like they might be our friend.

It turns out that it’s a bit less messy to work out $2S$.

  • $\frac{4k+2}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}$
  • $4k+2 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)$ [*]

Aside: let’s solve this two ways

First way: simultaneous equations. I don’t much care for this way, but it seems that some do.

Matching coefficients:

  • $[k^2]$: $0 = A + B + C$
  • $[k]: 4 = 3A + 2B + C$
  • $[1]: 2 = 2A$

This gives us $A=1$ immediately, with $B= 2$ and $C=-3$ close behind.

Alternatively, we can substitute any numbers we like into [*] and get:

  • $[k=0]$: $2 = A$
  • $[k=-1]: -2 = -B$, so $B=2$
  • $[k=-2]: -6=2C$, so $C=-3$, as before.

Back to the question at hand

So, $2S = \sum_{k=1}^{\infty} \left( \frac{1}{k} + \frac{2}{k+1} - \frac{3}{k+2}\right)$.

Let’s write out a few terms of that:

k          
1 $\frac{1}{1}$ $\frac{2}{2}$ $-\frac{3}{3}$    
2 . $\frac{1}{2}$ $\frac{2}{3}$ $-\frac{3}{4}$  
3 . . $\frac{1}{3}$ $\frac{2}{4}$ $-\frac{3}{5}$

It’s hopefully quite clear that the terms with a denominator of 3 sum to zero, and so will the terms with a denominator of 4, 5 and so on.

All we’re left with are the denominators of 1 (which sum to 1) and 2 (which sum to $\frac{3}{2}$), so $2S = \frac{5}{2}$ and $S = \frac{5}{4}$.

Alternatively

We could use generating functions. It’s massively overkill, but it’s a good excuse to use the technique. Let’s imagine that our sum is in fact a function of $x$, $S(x)= \sum_{k=1}^{\infty} \frac{2k+1}{k(k+1){k+2}}x^k$.

We can reuse our partial fractions and rewrite that as $2S(x) = \sum_{k=1}^{\infty} \left( \frac{x^k}{k} + \frac{2x^k}{k+1} - \frac{3x^k}{k+2}\right)$

And what we have there are three logarithmic series. Let’s examin them in turn:

  • $\sum_{k=1}^{\infty}\frac{x^k}{k} = \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \dots$

  • $\dots = -\ln(1-x)$.

  • $\sum_{k=1}^{\infty} \frac{2x^k}{k+1} = 2\left(\frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} + \dots \right)$

  • $\dots = -\frac{2}{x}\left(\ln(1-x)+x\right)$.

  • $\sum_{k=1}^{\infty} -\frac{3x^k}{k+2} = -3\left( \frac{x}{3} + \frac{x^2}{4} + \frac{x^3}{5} \dots \right)$

This one’s a bit trickier. I would start with $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$ and think “I want to get rid of the first couple of terms and divide by $x^2$”, giving me…

  • $\dots = 3\left( \frac{\ln(1-x)+x+\frac{x^2}{2}}{x^2}\right)$
  • or more nicely, $\frac{3}{2x^2}\left(2\ln(1-x)+2x + x^2\right)$

So, altogether, $2S(x) = \frac{3}{2x^2} \left(2\ln(1-x) + 2x + x^2\right) - \frac{2}{x}\left(\ln(1-x)+x\right) - \ln(1-x)$.

That screams “regroup!” at me. I’m going to turn the whole thing into a single fraction:

  • $2S(x) = \frac{(6\ln(1-x) + 6x + 3x^2) - (4x\ln(1-x)+4x) - 2x^2\ln(1-x)}{2x^2}$
  • $\dots = \frac{(6 - 4x - 2x^2)\ln(1-x) + 3x^2 + 2x}{2x^2}$

We’re interested in the value of this when $x=1$, although there’s a little technical difficulty: this is undefined when $x=1$, because $\ln(0)$ is not a number.

Fortunately, the factor in front of the logarithm also evaluater to zero, and $\lim_{z\to0} z \ln(z) = 0$ ((Prove this, if you’d like to.))

So, in the limit, we’re left with $2S(1) \to \frac{3 + 2}{2}$, so $S(1) \to \frac{5}{4}$ as before.

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I’m absolutely not saying that the generating functions way is easier, but the technique is an interesting and wide-ranging one. Try applying it to your favourite difficult sequences today!