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Check my work? (series)

  1. Nov 8, 2013 #1
    Find the interval of convergence for the power series [itex]\sum[/itex] [itex]\frac{x^n}{\sqrt{n}}[/itex]

    using the ratio test I get that the absolute value of x * the lim of square root of n over square root of n+1 = 0. so that being said i believe the interval of convergence is (-∞,∞) by the ratio test
     
    Last edited: Nov 8, 2013
  2. jcsd
  3. Nov 8, 2013 #2

    vanhees71

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    Think about your limit again! The convergence radius is
    [tex]\lim_{n \rightarrow \infty} \frac{a_n}{a_{n+1}}=\lim_{n \rightarrow \infty} \sqrt{\frac{n+1}{n}}=\cdots[/tex]
     
  4. Nov 8, 2013 #3
    Why? using the ratio test you get X^n+1 / sqrt(n+1) * the reciprocal of the original expression. So the x^n cancel out. leaving it in the form with n / n+1 square root
     
  5. Nov 8, 2013 #4
    also i notice you have n+1 in the denominator. Isnt it the numerator in the original formula? thats how it is in my book anyways
     
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