Find the sum of the infinite series ∑_(n=1)^(+∞)▒1/(k^2+5k+6)
Given series is-
∑k=1∞1k2+5k+6\sum_{k=1}^{\infty}\dfrac{1}{k^2+5k+6}∑k=1∞k2+5k+61
=∑k=1∞1(k+3)(k+2)=∑k=1∞(1(k+2)−1(k+3)=(13−14)+14−15)+15−16+.....=(13+14+15+.....)−(14+15+.....)=13=\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}=∑k=1∞(k+3)(k+2)1=∑k=1∞((k+2)1−(k+3)1=(31−41)+41−51)+51−61+.....=(31+41+51+.....)−(41+51+.....)=31
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