1
$\begingroup$

A fraction $\dfrac{a}{bc}$ can be split into $\dfrac{x}{b} + \dfrac{y}{c}$ by solving for $x$ and $y$ from $a = by + cx$.

Now, then a term $\dfrac{a}{cd}$ should be able to be split like this $$\frac{x}{bc} + \frac{y}{d}$$

But apparently that is not the case, as I have redone the decomposition for the expression $\dfrac{1}{k(k + 1)(k + 2)}$ several times now:

\begin{align*} \frac{1}{k(k + 1)(k + 2)} &= \frac{\overbrace{1}^a}{\underbrace{(k^2 + k)}_b\underbrace{(k + 2)}_c}\\ &= \frac{x}{(k^2 + k)} + \frac{y}{(k + 2)}&\\ &= \frac{x(k + 2)}{(k + 2)(k^2 + k)} + \frac{y(k^2 + k)}{(k + 2)(k^2 + k)} \end{align*}

Solving for $x$ and $y$:

\begin{align*} & 1 = x(k + 2) + y(k^2 + k)\\ & k := -2 \implies y = \frac{1}{2}\\ & 1 = x(k + 2) + \frac{1}{2}(k^2 + k)\\ & k := -1 \implies x = 1 \end{align*}

Thus:

$$\frac{1}{k(k + 1)(k + 2)} = \frac{1}{k(k + 1)} + \frac{\frac{1}{2}}{(k + 2)}\\$$

But that is not correct, as this example shows:

$$\frac{1}{3(3 + 1)}+ \frac{\frac{1}{2}}{3 + 2} = \frac{11}{60}$$

while

$$\frac{1}{3(3 + 1)(3 + 2)} = \frac{1}{60}$$

I hope I have not made some silly obvious arithmetic or algebraic mistake, but I have redone this several times, leading me to assume I am making a wrong assumption about how this works. So, is my assumption that there are $x$ and $y$ such that $\dfrac{a}{bcd} = \dfrac{x}{bc} + \dfrac{y}{d}$ wrong?

$\endgroup$
  • 1
    $\begingroup$ You should guess $\frac{ak+b}{k(k+1)}$ instead. $\endgroup$ – Chee Han Dec 12 '16 at 0:43
  • $\begingroup$ @Moo why is that? $\endgroup$ – user3578468 Dec 12 '16 at 0:46
  • $\begingroup$ @Moo ok, but why can't I make $k(k+1)$ to $k^2 + k$ and then have $\frac{a}{k^2 + k} + \frac{b}{k+2}$? Why do I have to have one fraction for each factor in the denominator? $\endgroup$ – user3578468 Dec 12 '16 at 0:55
  • $\begingroup$ @Moo ok, but when I do not make a fraction for each factor then I would be doing it like I did it in my question post, which lead to a wrong result. which would imply what I did was in principle correct and I just made some simple error OTHER than splitting the fraction the way I did. But I don't see any error, which in turn does not reduce my confusion. $\endgroup$ – user3578468 Dec 12 '16 at 1:03
  • $\begingroup$ @Moo why did you delete your comments? ... $\endgroup$ – user3578468 Dec 12 '16 at 1:19
1
$\begingroup$

You have three factors without multiplicity, so try

$$\frac{1}{k(k+1)(k+2)} \equiv \frac{A}{k}+\frac{B}{k+1}+\frac{C}{k+2}$$

Spoiler below

$$\frac{1}{k(k+1)(k+2)} \equiv \frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}$$

$\endgroup$
0
$\begingroup$

Your procedure is ok till $1 = x(k + 2) + y(k^2 + k)$.
Now in solving this you make the wrong assumption that $x$ is independent from $k$ (is constant for whichever $k$). The theory of partial fractions decomposition says that if the denominator of one the decomposition fractions is (or is kept) of degree 2, then the numerator will, in general, be of degree 1.
The answer above shows you the details.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.