# Prove $$\displaystyle{\frac{{{1}}}{{{4}{n}^{{2}}-{1}}}}={\frac{{{1}}}{{{\left({2}{n}+{1}\right)}{\left({2}{n}-{1}\right)}}}}={\frac{{{1}}}{{{2}{\left({2}{n}-{1}\right)}}}}-{\frac{{{1}}}{{{2}{\left({2}{n}+{1}\right)}}}}$$ Could you explain the operation in

Prove $\frac{1}{4{n}^{2}-1}=\frac{1}{\left(2n+1\right)\left(2n-1\right)}=\frac{1}{2\left(2n-1\right)}-\frac{1}{2\left(2n+1\right)}$
Could you explain the operation in the third step?
$\frac{1}{4{n}^{2}-1}=\frac{1}{\left(2n+1\right)\left(2n-1\right)}=\frac{1}{2\left(2n-1\right)}-\frac{1}{2\left(2n+1\right)}$
It comes from the sumation
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4{n}^{2}-1}$
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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memantangti17
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:
$\frac{1}{\left(2n+1\right)\left(2n-1\right)}=\frac{A}{2n+1}+\frac{B}{2n-1}$
Now multiply both sides by $\left(2n+1\right)\left(2n-1\right)$ and solve for A and B.