# Prove $\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$

Could you explain the operation in the third step?

$$\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$

It comes from the sumation $$\sum_{n=1}^\infty\frac1{4n^2-1}$$

I could just copy it to my homework, but I'd like to know how this conversion works. Thanks in advance.

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The motivation is to write $\dfrac1{(2n+1)(2n-1)}$ as $\dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$. Hence, we need $A$ and $B$ such that $$\dfrac1{(2n+1)(2n-1)} = \dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$$ Simplifying the right hand side, we get that $$\dfrac1{(2n+1)(2n-1)} = \dfrac{A(2n+1) + B(2n-1)}{(2n+1)(2n-1)} = \dfrac{2(A+B)n + (A-B)}{(2n+1)(2n-1)}$$ Hence, we need $2(A+B)n + (A-B) = 1$ for all $n$. Hence, matching the coefficient of $n$ and the constant term we get that $A-B = 1$ and $2(A+B) = 0$. This gives us $B = -\dfrac12$ and $A = \dfrac12$. Hence, $$\dfrac1{(2n+1)(2n-1)}=\dfrac12 \dfrac1{(2n-1)} - \dfrac12 \dfrac1{(2n+1)}$$

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Wow, just one seemingly insignificant step and so much is hidden in behind. Thanks for explaining. –  Grant Nov 8 '12 at 2:38

Instead of moving from the second expression to the third, try to move from the third to the second.

More generally, this technique is called "decomposing into partial fractions."

In this particular case, you might start by writing: $$\frac{1}{(2n+1)(2n-1)} = \frac{A}{2n+1} + \frac{B}{2n-1}$$

Now multiply both sides by $(2n+1)(2n-1)$ and solve for $A$ and $B$.

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I'll shed some light on another way to do this using a slightly different method with partial fractions. Note that the way Marvis did it is generally a more effective method for more difficult expressions (say with several variables where systems of equations are nice to work with). $$\frac{1}{{4n^2 - 1}} = \frac{1}{(2n+1)(2n-1)} = \frac{A}{2n+1} + \frac{B}{2n-1}$$

We can cross multiply to get

$$1 = A(2n-1) + B(2n+1)$$

To easily find the values of $A$ and $B$, it would be nice if we were able to get it into a single variable linear equation rather than having to deal with systems of equations. This is the same as saying that we want to try to find a $n$ such that $A(2n-1) = 0$ and $B(2n+1) = 0$.

$$2n-1 = 0 \implies 2n = 1 \implies n = \frac{1}{2}$$ $$2n + 1 = 0 \implies 2n = - 1 \implies n = -\frac{1}{2}$$

So, let's try both of those values now.

Plugging in $n = \frac{1}{2}$ yields

$$1 = 2B \iff B = \frac{1}{2}$$

Plugging in $n = -\frac{1}{2}$ yields

$$1 = -2A \iff A = -\frac{1}{2}$$

Hence, $$\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$

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