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

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
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.