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Hitaro9

  • one year ago

Series, how do I show (sqrt(k+1)-sqrt(k))/k converges?

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  1. hitaro9
    • one year ago
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    I imagine this is really simple, probably a comparison test, all the other problems in this section have been 1 or 2 steps, I'm just not seeing it.

  2. hitaro9
    • one year ago
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    \[\sum_{0}^{infinity}(\sqrt(k+1) -\sqrt(k))/k\]

  3. anonymous
    • one year ago
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    wait does it converge to 0 if it is so....i can give you an explanation

  4. ParthKohli
    • one year ago
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    The starting index should probably be 1.

  5. hitaro9
    • one year ago
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    Oh yeah, starting index is 1 my bad

  6. anonymous
    • one year ago
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    yeah,, parth is right

  7. hitaro9
    • one year ago
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    And uhh, the sequence goes to 0. If it didn't then the problem would be easy, would just say "lim =/= 0 therefore diverges" But the series doesn't converge to 0 I don't think?

  8. anonymous
    • one year ago
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    \[\large \sum_{k=1}^{\infty} \left(\frac{\sqrt{k+1} -\sqrt{k}}{k}\right)\]

  9. anonymous
    • one year ago
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    \[\sqrt{k-1/k} -1\] \[\sqrt{1-1/k} -1\] if k goes to infinity 1/k goes to zero then \[\sqrt{1}-1=0\]

  10. ParthKohli
    • one year ago
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    but you've got to show the convergence of the series

  11. hitaro9
    • one year ago
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    Yeah, it's not enough to show the sequence goes to 0, otherwise 1/n would converge.

  12. hitaro9
    • one year ago
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    Unless I'm missing something about what you're saying.

  13. anonymous
    • one year ago
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    Could you perhaps use a ratio test to show convergence?

  14. anonymous
    • one year ago
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    i think i should that on ... do i have to use the summation sign?

  15. anonymous
    • one year ago
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    oh wait... i got what ur saying now

  16. hitaro9
    • one year ago
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    Ratio test would end up with a bunch of n+1's around right? None of which would cancel nicely? Maybe you're right, I'll jot some notes real quick and see if anything is obvious

  17. anonymous
    • one year ago
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    \[\lim_{k\rightarrow \infty} \sqrt[n]{|a_n|} \ne 1\] The root test might work too.

  18. anonymous
    • one year ago
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    whoop, meant o write n instead of k*

  19. hitaro9
    • one year ago
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    Oh right didn't think of the root test. That might work out more nicely.

  20. hitaro9
    • one year ago
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    So you'd end up with some k^(k+1/2) type stuff, all of which would go to 1 right?

  21. hitaro9
    • one year ago
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    |dw:1443515996560:dw|Ugly drawing, but because the tops both go to 1, it'll go to 0/1, meaning it converges?

  22. hitaro9
    • one year ago
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    Is that right?

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