Series test.

**rushton** · May 20th 2011, 10:15 PM

Ok so I am a little stuck on this question, I need to test this series using only the standard comparison test.
$\sum_{n = 2}^\infty \frac{\left\vert{\sin n } \right\vert{} }{{n}^{2 }-1}$

I can't compare it to $\frac{1}{{n}^{ 2}}$ can I?

**Prove It** · May 20th 2011, 10:20 PM

Since $\displaystyle |\sin{n}| \leq 1$ , you can compare this series to $\displaystyle \sum_{n = 2}^{\infty}{\frac{1}{n^2 - 1}}$ . If you use partial fractions, you should find that this series is telescopic...

**rushton** · May 20th 2011, 10:29 PM

Thanks, also what about
$\sum_{n = 0}^\infty \frac{{n}^{3 } }{{n}^{ 4}+1 }$

**Prove It** · May 20th 2011, 10:34 PM

I would use the integral test, because I can see that $\displaystyle \int{\frac{x^3}{x^4 + 1}\,dx}$ is easily evaluated.

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{\frac{n^3}{n^4 + 1}} &\geq \int_0^{\infty}{\frac{x^3}{x^4 + 1}\,dx} \\ &= \frac{1}{4}\int_0^{\infty}{\frac{4x^3}{x^4 + 1}\,dx} \\ &= \frac{1}{4}\int_1^{\infty}{\frac{1}{u}\,du} \\ &= \frac{1}{4}\left[\lim_{\epsilon \to \infty}\left(\ln{\epsilon}\right) - \ln{1}\right] \\ & \to \infty\end{align*}$ .

So the series is divergent.

**rushton** · May 20th 2011, 10:46 PM

We can't unfortunately. Only the comparison test and the arithmetic, geometric, harmonic series and $\frac{1}{ {n}^{k } }$ series.

I let n= ${2}^{ m}$ and got $\frac{1}{{2}^{ m}+\frac{1}{ {2}^{ 3m} }} < \frac{1}{{2}^{ m} }$

which is geometric and converges. But I'm not at all confident with that.

**rushton** · May 20th 2011, 10:48 PM

Ok that gives me something to go off I will try and figure out to see if I can show it is divergent somehow

**rushton** · May 20th 2011, 11:10 PM

Ok I think I got it.

$\sum_{n = 1}^\infty \frac{{n}^{3 } }{ {n}^{4 }+1 } \geqslant \sum_{n = 1}^\infty \frac{1}{n+1}$ which diverges so

$\sum_{n = 0}^\infty \frac{{n}^{3 } }{ {n}^{4 }+1 }$ must diverge.

Does this sound right?

**Prove It** · May 20th 2011, 11:12 PM

Ho do you know that $\displaystyle \frac{n^3}{n^4 + 1} \geq \frac{1}{n + 1}$ ?

**rushton** · May 20th 2011, 11:21 PM

For $n\geqslant 1 \frac{{n}^{3 } }{{n}^{4 } +1 } \geqslant \frac{{n}^{3 } }{{n}^{4 } +{n}^{ 3} } = \frac{1}{ n+1}$

**Prove It** · May 21st 2011, 05:55 AM

OK I agree with your comparison then

**rushton** · May 24th 2011, 10:25 PM

Thanks a lot you were a real help

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