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Math Help - convergence divergence question quick and easy :D

  1. #1
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    convergence divergence question quick and easy :D

    when taking the summation of 1/n from 1 to infinity,

    what is the difference between saying that it diverges or converges to 0?

    I am kind of puzzled by this because, for example the summation from 1 to infinity of ln(2x+3)/x , what does this do?

    does it converge to 0 after using l'hopitals rule, or does it follow the harmonic series rule and diverge?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by p00ndawg View Post
    when taking the summation of 1/n from 1 to infinity,

    what is the difference between saying that it diverges or converges to 0?

    I am kind of puzzled by this because, for example the summation from 1 to infinity of ln(2x+3)/x , what does this do?

    does it converge to 0 after using l'hopitals rule, or does it follow the harmonic series rule and diverge?
    No...it does not ...the first one diverges because it is a divergent P-series..the second one diverges by a comparison to \frac{\ln(x)}{x} which diverges by the integral test...if you need A LOT of help look here http://www.mathhelpforum.com/math-he...-tutorial.html
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    Quote Originally Posted by Mathstud28 View Post
    No...it does not ...the first one diverges because it is a divergent P-series..the second one diverges by a comparison to \frac{\ln(x)}{x} which diverges by the integral test...if you need A LOT of help look here http://www.mathhelpforum.com/math-he...-tutorial.html
    naw I dont need a lot of help, but I got a test tomorrow and just have a few holes i needed some filling on.

    you say 1/n is divergent because of the p series, but isnt the series for it being greater than or less than 1? if its a big proof just reply with because I said so, ill understand.

    but, so are you saying that if a question or function reduces through some kind of comparison or test, like the integral or nth term test, then if it does reduce to 1/n it would actually converge instead of diverge?

    oh and im sorry the equation is ln(2x^3 +1)/x
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    Quote Originally Posted by p00ndawg View Post
    when taking the summation of 1/n from 1 to infinity,

    what is the difference between saying that it diverges or converges to 0?

    I am kind of puzzled by this because, for example the summation from 1 to infinity of ln(2x+3)/x , what does this do?

    does it converge to 0 after using l'hopitals rule, or does it follow the harmonic series rule and diverge?
    To answer both your questions:

    The difference between a series diverging or converging to zero is a big difference. When a series diverges, that means the series has no sum. If a series converges to zero, that means that the sum of the series exists and is equal to zero.

    The series \sum\frac{\ln(2x+3)}{x} diverges; although L'Hospital's Rule can tell you that the limit of the terms is zero, this does not imply that the series converges. If the limit of the terms (as n approaches infinity) is not zero, then you can conclude the series diverges.

    You show that \sum\frac{\ln(2x+3)}{x} diverges by comparison with the harmonic series.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by p00ndawg View Post
    naw I dont need a lot of help, but I got a test tomorrow and just have a few holes i needed some filling on.

    you say 1/n is divergent because of the p series, but isnt the series for it being greater than or less than 1? if its a big proof just reply with because I said so, ill understand.

    but, so are you saying that if a question or function reduces through some kind of comparison or test, like the integral or nth term test, then if it does reduce to 1/n it would actually converge instead of diverge?
    another way is this...since it decreases as n gets bigger and all the terms are positive we can apply the integral test \int_1^{\infty}\frac{1}{n}dn=\ln(n)\bigg|_1^{\inft  y}=\infty-0 therefore it is divergent....and if a p-series has exponents 0<x<1 it diverges
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    Quote Originally Posted by Mathstud28 View Post
    another way is this...since it decreases as n gets bigger and all the terms are positive we can apply the integral test \int_1^{\infty}\frac{1}{n}dn=\ln(n)\bigg_|1^{\inft  y}=\infty-0 therefore it is divergent....and if a p-series has exponents 0<x<1 it diverges
    hey sorry about this but the equation was  ln(2x^3+1)/x
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    Quote Originally Posted by icemanfan View Post
    To answer both your questions:

    The difference between a series diverging or converging to zero is a big difference. When a series diverges, that means the series has no sum. If a series converges to zero, that means that the sum of the series exists and is equal to zero.

    The series \sum\frac{\ln(2x+3)}{x} diverges; although L'Hospital's Rule can tell you that the limit of the terms is zero, this does not imply that the series converges. If the limit of the terms (as n approaches infinity) is not zero, then you can conclude the series diverges.

    You show that \sum\frac{\ln(2x+3)}{x} diverges by comparison with the harmonic series.
    does it change if the equation is ln(2x^3+3/x) ?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by p00ndawg View Post
    hey sorry about this but the equation was  ln(2x^3+1)/x
    Quote Originally Posted by p00ndawg View Post
    does it change if the equation is ln(2x^3+3/x) ?
    The first one still diverges by comparison to that \frac{\ln(x)}{x} and the second one diverges since \lim_{x\to\infty}\ln\bigg(2x^2+\frac{3}{x}\bigg)=\  infty\ne{0}...which is the n-th term test
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  9. #9
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    Quote Originally Posted by p00ndawg View Post
    does it change if the equation is ln(2x^3+3/x) ?
    That series diverges even faster. And in that case, you can use the nth term test to prove it.
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