Why does this partial fraction decomposition work, even with division by $0$?

Question

Here is how partial fraction decomposition works. First, take a fraction like $\frac{1}{n(n+1)}$. You can express this fraction as the sum $\frac{A}n + \frac{B}{n+1}$ for some constants $A$ and $B$. Thus, you have the equation $\frac{1}{n(n+1)} = \frac{A}n + \frac{B}{n+1}$. Now, multiply both sides by $n(n+1)$ to get $1=A(n+1)+B(n)$. However, note that the expression $\frac{x}{x} = 1$ ONLY when $x \ne 0$ because division by $0$ is forbidden. Thus, $1=A(n+1)+B(n)$ only if $n \ne 0,-1$ because these are the values that cause division by $0$. However, many people plug in $n=0,-1$ to solve for $A,B$ even though that is an illegal move, and it works.

Plugging in $n=0$, we get $1=A$. Plugging in $n=-1$, we get $1 =-B$, thus $B=-1$. So, we have $\frac{1}{n(n+1)} = \frac{A}n + \frac{B}{n+1} = \frac{1}n - \frac{1}{n+1}$. This is true.

However, I don't understand why it works, even though you used an illegal move?

Answer 1

This is a good question, and it's quite subtle and almost no one mentions it; the reason is continuity. Let $P:\Bbb{R}\to\Bbb{R}$ be a polynomial function, $\{a_1,\dots, a_k\}$ all of its roots, and let $f:\Bbb{R}\to\Bbb{R}$ be a continuous function. Suppose you know that for all $x\in\Bbb{R}\setminus\{a_1,\dots, a_k\}$ \begin{align} \frac{1}{P(x)}&=\frac{f(x)}{P(x)} \end{align} Then, for these values of $x$, it follows that $1=f(x)$. Now since $f$ is a continuous function, the same equality holds even if $x\in\{a_1,\dots, a_k\}$. Therefore, for every $x\in\Bbb{R}$ without exception, we have the equality $1=f(x)$.

In your case, we're applying this reasoning to $P(n)= n(n+1)$ and $f(n)=A(n+1)+Bn$.

---

More generally, we can make the following observation:

Let $P:\Bbb{R}\to\Bbb{R}$ be a polynomial function, $f,g:\Bbb{R}\to\Bbb{R}$ be continuous functions, and suppose that for every $x\in\Bbb{R}$ which is not a zero of $P$, we have \begin{align} \frac{g(x)}{P(x)}&= \frac{f(x)}{P(x)} \end{align} Then, for every $x\in\Bbb{R}$ (without exception) we have $f(x)=g(x)$.

Answer 2

$A(n+1)+Bn$ is a polynomial in $n$, as is $1$. You want values of $A$ and $B$ that make these two polynomials equal for all real numbers except $0$ and $-1$. But two polynomials that agree at infinitely many real numbers are necessarily identically equal and will therefore agree at all real numbers. In particular, $A(n+1)+Bn=1$ will hold at $n=0$ and $n=-1$ if it holds at all other integers, so we can use $n=0$ and $n=-1$ as shortcuts to finding the correct values of $A$ and $B$.

Answer 3

Let's think of these expressions as functions on subsets of the real numbers. i.e. $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$ for $x \neq -1, 0$. Then $1=A(x+1)+B(x)$ for $x \neq -1, 0$.

You're correct to note that we can't plug in $-1$ and $0$ straight away because the functions aren't defined there. However, since the functions are defined on a set with $-1$ and $0$ as limit points we can take the limit as $x$ tends to the forbidden points. Since the functions $1$ and $A(x+1)+B(x)$ are equal everywhere else they have the same limit.

Since the functions are continuous the limit is exactly the same as plugging in the forbidden value.