How can I make sense of this basic algebraic manipulation backwards?

Question

$$\frac{1}{n^2-1} = \frac12\left(\frac1{n-1}-\frac1{n+1}\right)$$

It is easy for me to understand this algebraic from right to left ←, but I struggle to find a reasonable way to work from left to right.

Regarding expected difficulty, the level of algebraic manipulated is here meant for a Calculus II course.

Thanks all.

Answer 1

Think about passing to a common denominator, in reverse: whenever we see a fraction of the form ${a\over b\cdot c}$, we should expect it to equal ${x\over b}+{y\over c}$ for some reasonable $x, y$.

In fact, once we get this idea, it's not hard to see that this will happen whenever $x$ and $y$ satisfy $$xc+yb=a.$$ Of course, this is sort of "right-to-left" as far as the OP goes, but it's motivated by the left-to-right direction. OK, so let's look at the specific case. Here, $a=1$, $b=n-1$, $c=n+1$ (since $n^2-1=(n-1)(n+1)$). So we want $x$ and $y$ to satisfy $$x(n+1)+y(n-1)=1.$$ This might look messy, but we have one more trick: we want $x$ and $y$ to be just numbers, not expressions in terms of $n$. (Why? Well, just because if we could, that'd make our lives easier. :P) If this is going to work, though, then the $n$s on the left hand side have to cancel out - so we have to have $$x=-y.$$ Aha! Now that simplifies to get $$x+(-x)(-1)=1,$$ that is, $$x={1\over 2}$$ and so $$y=-{1\over 2}.$$

Answer 2

Write $$R = \frac{1}{n^2 - 1} = \frac{1}{(n - 1)(n + 1)}$$

Now, we wish to decompose $R$ into something with denominator $(n - 1)$ and denominator $(n + 1)$

The standard way to do this is by a method knows as "partial fractions". We assign $R$ to

$$R = \frac{A}{n - 1} + \frac{B}{n + 1} = \frac{1}{(n - 1)(n + 1)}$$

Cross multiplying,

$$R = \frac{A (n + 1) + B (n - 1)}{(n - 1)(n + 1)} = \frac{1}{(n - 1)(n + 1)} $$

Since the denominators are equal, we can simply equate the numerators and see what values of A, B we need.

$$ A(n + 1) + B(n - 1) = 1 + 0n \\ (A - B) + n(A + B) = 1 + 0n $$

Clearly, we have two inequalities (by assigining the coefficients of the constant term and $n$ to each other.

$$ A - B = 1 \\ A + B = 0 $$

Substituting $B = -A$ in the first equation,

$$ A - (-A) = 1 \\ 2A = 1 \\ A = \frac{1}{2} $$

Substituting $A = \frac{1}{2}$ in $A + B = 0$ gives $B = \frac{-1}{2}$

Hence, $$R = \frac{\frac{1}{2}}{n - 1} + \frac{\frac{-1}{2}}{n + 1} \\ = \frac{1}{2}\left(\frac{1}{n - 1} - \frac{1}{n + 1} \right)$$