1. ## series?convergent

$\displaystyle \sum_{n=0}^{\infty}3+sin(n)/n^3+2$
how would i do this? all i know is the sequence of partial sums is a finite vaule , 0? correct wouldnt that mean that my series will be convergent?

2. Originally Posted by zangestu888
$\displaystyle \sum_{n=0}^{\infty}3+sin(n)/n^3+2$
how would i do this? all i know is the sequence of partial sums is a finite vaule , 0? correct wouldnt that mean that my series will be convergent?
If you mean

$\displaystyle \frac{3+\sin(n)}{n^3+2}$

Then by the comparison test

$\displaystyle \left| \frac{3+\sin(n)}{n^3+2} \right|< \frac{4}{n^3}$

The latter series is convergent by p-series or the integral test

3. Hello, zangestu888!

$\displaystyle S \;=\;\sum_{n=0}^{\infty}\frac{3+\sin(n)}{n^3+2}$
I would use the Comparison Test.

Since $\displaystyle \sin(n) \:\leq\: 1$, then: .$\displaystyle 3 + \sin(n) \:\leq 4$ .[1]

Since $\displaystyle n^3 + 2 \:>\: n^3$, then: .$\displaystyle \frac{1}{n^3+2} \:< \:\frac{1}{n^3}$ .[2]

Multiply [1] and [2]: .$\displaystyle \frac{3+\sin(n)}{n^3+2} \:< \frac{4}{n^3}$

Take sums: .$\displaystyle S \;=\;\sum^{\infty}_{n=1}\frac{3+\sin(n)}{n^3+2} \;< \;\sum^{\infty}_{n=1}\frac{4}{n^3} \quad\Leftarrow\:\text{ a convergent }p\text{-series}$

Therefore, $\displaystyle S$ converges.

Too slow again . . . $\displaystyle \emptyset$ beat me to it.
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