series?convergent

**zangestu888** · March 8th 2009, 06:51 PM

$\sum_{n=0}^{\infty}3+sin(n)/n^3+2$
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?

**TheEmptySet** · March 8th 2009, 07:05 PM

Originally Posted by zangestu888

$\sum_{n=0}^{\infty}3+sin(n)/n^3+2$
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

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

Then by the comparison test

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

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

**Soroban** · March 8th 2009, 07:08 PM

Hello, zangestu888!

$S \;=\;\sum_{n=0}^{\infty}\frac{3+\sin(n)}{n^3+2}$

I would use the Comparison Test.

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

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

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

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

Therefore, $S$ converges.

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

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