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

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The series in question is $\displaystyle\sum_{n=1}^{\infty}\dfrac{1-2\sin n}{3n^2+2n+4}$. It is established that the absolute value of the series can be bounded by $\dfrac{3}{3n^2+2n+4}$. Since the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{3}{3n^2+2n+4}$ is convergent, it follows that the original series is also absolutely convergent. Therefore, the conclusion is that the series is convergent.
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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.
 
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