Convergent/Divergent series

I really don't understand convergent/divergent series. I understand that you first use the ratio test, and then if L=1...try and find something else?

First thing's first. I reeeally don't understand when you find L, and simplify it (well i try my best), how do you then say if its greater than/less than 1?? What do you use. I'm being well stupid about this.

Say i've got this series:

\sum_{n=1}^\infty (2n^{-2} + 3n^{-3})

And I've got to:

L = \lim_{x\to \infty} | \frac{2n^{-1} + 3n^{-2}} {2n^{-2} + 3n^{-3}} \frac{1} {2n^{-2} + 3n^{-3}} |

L = \lim_{x\to \infty} | \frac{2n^{-1} + 3n^{-2}} {4n^{-4} + 9n^{-6}} |

Reply 1

Hopple

18

I'm not sure what you've done there :s-smilie:

Presumably you're trying to get

\lim_{n\to \infty} | \frac{a_{n+1}}{a_n}|

?

If so, I don't think that'll get you anywhere. I think you're supposed to notice that

\sum_{n=1}^\infty \frac{1}{n^a}

converges for a>1 and use that. If not, that was a spoiler for later on in your course :tongue:

Reply 2

ztibor

10

Original post by Mother_Russia

I really don't understand convergent/divergent series. I understand that you first use the ratio test, and then if L=1...try and find something else?

First thing's first. I reeeally don't understand when you find L, and simplify it (well i try my best), how do you then say if its greater than/less than 1?? What do you use. I'm being well stupid about this.

Say i've got this series:

\sum_{n=1}^\infty (2n^{-2} + 3n^{-3})

And I've got to:

L = \lim_{x \to \infty} | \frac{2n^{-1} + 3n^{-2}} {2n^{-2} + 3n^{-3}} \frac{1} {2n^{-2} + 3n^{-3}} |

L = \lim_{x\to \infty} | \frac{2n^{-1} + 3n^{-2}} {4n^{-4} + 9n^{-6}} |

\displaystyle \sum^{\infty}_{n=1} a_n

convergent if there is a
number of q so that

\frac{a_{n+1}}{a_n} \le q <1

But better if you use the limit L<1 criteria, that is that q number will be L limit
and

L=lim_{n \to \infty}\frac{a_{n+1}}{a_n}

There is a theorem: given the following series with positive terms

\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^{\alpha}}

convergent if

\alpha \ge 1

divergent otherwise

(edited 13 years ago)

Reply 3

davros

16

Original post by Mother_Russia

I really don't understand convergent/divergent series. I understand that you first use the ratio test, and then if L=1...try and find something else?

First thing's first. I reeeally don't understand when you find L, and simplify it (well i try my best), how do you then say if its greater than/less than 1?? What do you use. I'm being well stupid about this.

Say i've got this series:

\sum_{n=1}^\infty (2n^{-2} + 3n^{-3})

And I've got to:

L = \lim_{x\to \infty} | \frac{2n^{-1} + 3n^{-2}} {2n^{-2} + 3n^{-3}} \frac{1} {2n^{-2} + 3n^{-3}} |

L = \lim_{x\to \infty} | \frac{2n^{-1} + 3n^{-2}} {4n^{-4} + 9n^{-6}} |

Have you been told specifically to use the ratio test on this question? Because the comparison test will get you result much more quickly unless you're being asked to practise the ratio test in different cases!

Convergent/Divergent series

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