- Thread starter
- #1

- Jan 29, 2012

- 661

Here is a link to the question:

Is this series convergent? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter Fernando Revilla
- Start date

- Thread starter
- #1

- Jan 29, 2012

- 661

Here is a link to the question:

Is this series convergent? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter
- #2

- Jan 29, 2012

- 661

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.