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

$a_n=\frac{2n-5}{n(n+1)}$
We can see that to the subsequent terms' numerator, $2$ is added and the denominator is derived by the formula $n(n+1)$ with $a_1=\frac{-3}{2}$, so $a_n=\frac{2n-5}{n(n+1)}$ for all $n$ natural numbers.