Answer to Question #189975 in Calculus for Moel Tariburu

Question

Find the sum of the infinite series ∑_(n=1)^(+∞)▒1/(k^2+5k+6)

Answer 1

Given series is-

$\sum_{k=1}^{\infty}\dfrac{1}{k^2+5k+6}$

$=\sum_{k=1}^{\infty} \dfrac{1}{(k+3)(k+2)} \\[9pt]=\sum_{k=1}^{\infty} (\dfrac{1}{(k+2)}-\dfrac{1}{(k+3)} \\[9pt]=(\dfrac{1}{3}-\dfrac{1}{4})+\dfrac{1}{4}-\dfrac{1}{5})+\dfrac{1}{5}-\dfrac{1}{6}+..... \\[9pt] =(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+.....)-(\dfrac{1}{4}+\dfrac{1}{5}+.....) \\[9pt]=\dfrac{1}{3}$