An awkward sum
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:
 $\sum_{k=1}^{\infty}\frac{x^k}{k} = \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{3} + \dots$

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

$\dots = \frac{2}{x}\left(\ln(1x)+x\right)$.
 $\sum_{k=1}^{\infty} \frac{3x^k}{k+2} = 3\left( \frac{x}{3} + \frac{x^2}{4} + \frac{x^3}{5} \dots \right)$
This one’s a bit trickier. I would start with $\ln(1x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$ and think “I want to get rid of the first couple of terms and divide by $x^2$”, giving me…
 $\dots = 3\left( \frac{\ln(1x)+x+\frac{x^2}{2}}{x^2}\right)$
 or more nicely, $\frac{3}{2x^2}\left(2\ln(1x)+2x + x^2\right)$
So, altogether, $2S(x) = \frac{3}{2x^2} \left(2\ln(1x) + 2x + x^2\right)  \frac{2}{x}\left(\ln(1x)+x\right)  \ln(1x)$.
That screams “regroup!” at me. I’m going to turn the whole thing into a single fraction:
 $2S(x) = \frac{(6\ln(1x) + 6x + 3x^2)  (4x\ln(1x)+4x)  2x^2\ln(1x)}{2x^2}$
 $\dots = \frac{(6  4x  2x^2)\ln(1x) + 3x^2 + 2x}{2x^2}$
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 wideranging one. Try applying it to your favourite difficult sequences today!