Answer
$\dfrac{1}{6}$
Work Step by Step
Given: $\Sigma_{n=3}^{\infty}\dfrac{1}{(2n-3)(2n-1)}$
Re-write the given series as: $\Sigma_{n=3}^{\infty}\dfrac{1}{(2n-3)(2n-1)}=\frac{1}{2}\Sigma_{n=3}^{\infty}[\dfrac{1}{2n-3}-\dfrac{1}{2n-1}]$
Apply limits, we get:
$\frac{1}{2}\Sigma_{n=3}^{\infty}[\dfrac{1}{2n-3}-\dfrac{1}{2n-1}]=\lim\limits_{m \to \infty}\frac{1}{2}\Sigma_{n=3}^{m}[\dfrac{1}{2n-3}-\dfrac{1}{2n-1}]$
$\lim\limits_{m \to \infty}\frac{1}{2}\Sigma_{n=3}^{m}[\dfrac{1}{2n-3}-\dfrac{1}{2n-1}]=\lim\limits_{m \to \infty}\frac{1}{2}\Sigma_{n=3}^{m}[\dfrac{1}{3}-\dfrac{1}{2m-1}]$
After simplifications, we get
$\lim\limits_{m \to \infty}\frac{1}{2}\Sigma_{n=3}^{m}[\dfrac{1}{3}-\dfrac{1}{2m-1}]=\dfrac{1}{6}$
