## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{6}$
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 $\lim\limits_{m \to \infty}\frac{1}{2}\Sigma_{n=3}^{m}[\dfrac{1}{3}-\dfrac{1}{2m-1}]=\dfrac{1}{6}$