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El_Arrow

  • one year ago

determine whether the series is convergent or divergent

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  1. El_Arrow
    • one year ago
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    \[\sum_{n=1}^{\infty} ((1/n) + (1/n+1))\]

  2. El_Arrow
    • one year ago
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    @SolomonZelman

  3. El_Arrow
    • one year ago
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    @campbell_st

  4. anonymous
    • one year ago
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    Well, the easiest test for convergence/divergence of a series is to take the limit of n approaches infinity.

  5. anonymous
    • one year ago
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    Given a series, \[\sum_{n=1}^{\infty} a_{n}\]. If \[\lim_{n \rightarrow \infty} a_{n} = 0 \], it is convergent. Otherwise, it is divergent.

  6. El_Arrow
    • one year ago
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    so you use the integral test for this one or no? @math1234

  7. El_Arrow
    • one year ago
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    |dw:1436734133011:dw|

  8. El_Arrow
    • one year ago
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    like that?

  9. anonymous
    • one year ago
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    Correct.

  10. anonymous
    • one year ago
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    What is the limit in this case?

  11. El_Arrow
    • one year ago
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    from t to 1 @math1234

  12. El_Arrow
    • one year ago
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    but how do you find the limit of that?

  13. anonymous
    • one year ago
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    \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } + \frac{ 1 }{ n + 1}\]

  14. anonymous
    • one year ago
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    The limit of the first term is 0 right? Same as the second term.

  15. El_Arrow
    • one year ago
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    so it completely diverges?

  16. zzr0ck3r
    • one year ago
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    That is not what the theorem says, it says that if the limit DOES not converge to 0, then the series does not converge. It does NOT say that if the limit converges to 0 then the series converges. Every horse has four legs, but not every four legged animal is a horse

  17. zzr0ck3r
    • one year ago
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    integral test is a good idea

  18. anonymous
    • one year ago
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    this does not converge, obviously:$$\frac1{n+1}\ge 0\implies \frac1{n}\le\frac1{n}+\frac1{n+1}\implies \sum_{n=1}^N\frac1n\le \sum_{n=1}^N\left(\frac1n+\frac1{n+1}\right)$$

  19. anonymous
    • one year ago
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    and yet we know that \(\sum_{n=1}^\infty\frac1n=\lim_{N\to\infty}\sum_{n=1}^N\frac1n\to\infty\) as the harmonic series diverges, and surely something bigger than the harmonic series diverges, too

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