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anonymous

  • one year ago

Series exercie

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  1. anonymous
    • one year ago
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    \[\sum_{1}^{\infty} \frac{ n }{ n ^{2} +1 }\]

  2. anonymous
    • one year ago
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    @Owlcoffee

  3. Owlcoffee
    • one year ago
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    Okay, what's the objective?

  4. anonymous
    • one year ago
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    to find if it is convergent,conditonally convergent or divergent

  5. anonymous
    • one year ago
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    @ayeshaafzal221 helped me yesterday with it,I just wanna see another way of solving it

  6. anonymous
    • one year ago
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    Help?

  7. mathmate
    • one year ago
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    Hint: use the integral test.

  8. anonymous
    • one year ago
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    I got it like this: \[\lim_{n \rightarrow \infty} \frac{ \frac{ n }{ n ^{2} } }{ \frac{ n ^{2} }{ n ^{2} +1}} = \lim_{n \rightarrow \infty} \frac{ n }{ 2 } =\frac{ \infty }{ 2 }=0\]

  9. anonymous
    • one year ago
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    Is my solution right?

  10. ganeshie8
    • one year ago
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    You could also use the comparison test : for \(n\ge 1\), we have \(n^2+1 \le n^2+n^2 = 2n^2\) \(\implies \dfrac{1}{n^2+1}\ge \dfrac{1}{2n^2}\) \(\implies \dfrac{n}{n^2+1}\ge \dfrac{n}{2n^2} = \dfrac{1}{2n}\) Since \(\sum \dfrac{1}{2n}\) doesn't converge as this is a harmonic series, the series \(\sum \dfrac{n}{n^2+1}\) doesn't converge by comparison test

  11. anonymous
    • one year ago
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    So my solution is not right?

  12. ganeshie8
    • one year ago
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    I would give you 1/10 for the attempt because I don't really get what test you have used..

  13. ganeshie8
    • one year ago
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    Also why do you think \(\dfrac{\infty}{2}\) is anything close to \(0\) ?

  14. mathmate
    • one year ago
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    \(\large\frac{n}{n^2+1}\ne\frac{n/n^2}{n^2/n^2+1}\) so it is not the equivalent series, in any case \(\inf/2=\inf\)

  15. anonymous
    • one year ago
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    i meant infinite,not 0,sorry

  16. ganeshie8
    • one year ago
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    do you know what does it mean to 'converge' or 'diverge' ?

  17. anonymous
    • one year ago
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    Nope,I know if it dose not equal 0,it means it is divergent,if it is 0,then it's complicated

  18. anonymous
    • one year ago
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    does*

  19. ganeshie8
    • one year ago
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    before trying the convergence tests, you need to know the meaning of terms "convergent sreies" and "divergent series"

  20. anonymous
    • one year ago
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    can I use L'Hospital for this exercise?

  21. anonymous
    • one year ago
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    I get 1/2 if I use L'Hospital and it means it is divergent,right?

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