Question #189975

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


1
Expert's answer
2021-05-12T03:04:27-0400

Given series is-


k=11k2+5k+6\sum_{k=1}^{\infty}\dfrac{1}{k^2+5k+6}


=k=11(k+3)(k+2)=k=1(1(k+2)1(k+3)=(1314)+1415)+1516+.....=(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}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS