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Jenny's question at Yahoo! Answers (Convergence of a series)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Jenny,

Your series is $\displaystyle\sum_{n=1}^{\infty}\dfrac{1-2\sin n}{3n^2+2n+4}$, then $$\left|1-2\sin n\right|\leq \left|1+(-2\sin n)\right|\leq 1+2\;\left|\sin n\right|\leq 1+2=3$$ As a consequence, $$\left|\dfrac{1-2\sin n}{3n^2+2n+4}\right|\leq \dfrac{3}{3n^2+2n+4}$$ But easily proved, the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{3}{3n^2+2n+4}$ is convergent, hence the given series is absolutely convergent.