$\displaystyle \sum_{n=0}^{\infty}3+sin(n)/n^3+2$
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?
Hello, zangestu888!
I would use the Comparison Test.$\displaystyle S \;=\;\sum_{n=0}^{\infty}\frac{3+\sin(n)}{n^3+2}$
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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