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Math2400

  • one year ago

radius of convergence and the interval of convergence for the given power series

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  1. Math2400
    • one year ago
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    \[\sum_{n=1}^{\infty} \frac{ n ^{2}x ^{2n} }{ n! }\]

  2. Math2400
    • one year ago
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    my radius is 1. And it think the interval is (-1,1) Can anyone verify if that's right?

  3. Zarkon
    • one year ago
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    what test did you use?

  4. Math2400
    • one year ago
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    ratio

  5. Zarkon
    • one year ago
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    do it again

  6. Math2400
    • one year ago
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    ok

  7. Math2400
    • one year ago
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    wait can the radius be infinity??

  8. Zarkon
    • one year ago
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    yes

  9. Math2400
    • one year ago
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    ok..and the interval is (-infin, infin) ?

  10. Zarkon
    • one year ago
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    yes

  11. Math2400
    • one year ago
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    lol kk thanks. that's what i got first but i wasn't sure if that was possible haha

  12. xapproachesinfinity
    • one year ago
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    i got oo as well :)

  13. Math2400
    • one year ago
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    perfect thanks!! :)

  14. xapproachesinfinity
    • one year ago
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    limit (x^2 . (n+1/n)^2/(n+1))=0

  15. Math2400
    • one year ago
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    how about this one if u don't mind: \[\sum_{n=1}^{\infty} \frac{ n^{e} }{ e ^{n} }\] i'm pretty sure the radius is 1 right?

  16. xapproachesinfinity
    • one year ago
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    that says for any value of x rho is always zero

  17. Math2400
    • one year ago
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    and then interval is (2,4) or maybe (2,4] not sure, i suck at these

  18. xapproachesinfinity
    • one year ago
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    i would do root test with that one

  19. xapproachesinfinity
    • one year ago
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    did you try ratio?

  20. Zarkon
    • one year ago
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    there is no x in your last problem

  21. Math2400
    • one year ago
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    yes(: I got the radius to be 1 and then i know the interval is (2,4) but I'm not sure

  22. xapproachesinfinity
    • one year ago
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    eh hold on you are asking for convergence because there is not x

  23. xapproachesinfinity
    • one year ago
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    you need x

  24. Math2400
    • one year ago
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    ohh wait sorry i posted the wrong one. That one is a converging series haha. I meant to post this one: \[\sum_{n=1}^{\infty} \frac{ (x-3)^n }{ \sqrt{n} }\]

  25. xapproachesinfinity
    • one year ago
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    oh ok i would go with root test with that too limit n^1/n=1 rho=|x-3|<1 so raduis is 1 -1<x-3<1 ===> 2<x<4

  26. xapproachesinfinity
    • one year ago
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    i guess you got this :)

  27. Math2400
    • one year ago
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    ok so radius is 1 and then did u check the end points?

  28. xapproachesinfinity
    • one year ago
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    No i didn't x=4 you have series 1/n^1/2 diverges x=2 (-1)^n/rootn i believe this converge by alternating series test

  29. Math2400
    • one year ago
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    so would it be (2,4] or just (2,4) that's what I'm confused on

  30. xapproachesinfinity
    • one year ago
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    no converge on 2 not on 4 so [2,4) otherwise 2<=x<4

  31. Math2400
    • one year ago
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    ohh haha kk got it :) Thank you so much!

  32. xapproachesinfinity
    • one year ago
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    no problem :)

  33. xapproachesinfinity
    • one year ago
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    actually you seem to me good at this, just few more practice and you master it

  34. Math2400
    • one year ago
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    haha thank you, means a lot @xapproachesinfinity

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