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Numbers for which it coverges

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the positive numbers r for which [tex]\sum \frac{r^k}{k^r}[/tex] converges.

    2. Relevant equations



    3. The attempt at a solution

    What method do I use for this?

    [tex]\frac {a_{k+1}}{a_k} = \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}
    = r \cdot (\frac{1}{k})^{\frac{r}{k}} [/tex]
     
  2. jcsd
  3. Mar 23, 2010 #2

    Mark44

    Staff: Mentor

    The second and third expressions aren't equal...
     
  4. Mar 23, 2010 #3
    Ok so I have [tex] \frac {a_{k+1}}{a_k} = \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}
    = r \cdot \frac{k^r}{(k+1)^r } = r \cdot (\frac{k}{(k+1)})^r[/tex] ?
     
  5. Mar 23, 2010 #4
    then [tex]r \cdot (1 - \frac{1}{k+1})^r \to r \cdot 1^r [/tex] ?
     
  6. Mar 23, 2010 #5
    then r needs to be < 1?
     
  7. Mar 23, 2010 #6

    Mark44

    Staff: Mentor

    What if r = -200? That's less than 1.
     
  8. Mar 23, 2010 #7
    I thought if the ratio is < 1 then it converges?
     
  9. Mar 23, 2010 #8

    Mark44

    Staff: Mentor

    The ratio test is usually given in terms of absolute values. IOW,
    [tex]\lim_n \frac{|a_{n + 1}|}{|a_n|}[/tex]

    For this problem it's given that r is a positive number, so since the ratio of those terms is less than 1, the series converges.
     
  10. Mar 24, 2010 #9
    So how do I know the ratio is less than 1?
     
  11. Mar 24, 2010 #10

    Mark44

    Staff: Mentor

    I thought you already knew that. What I thought you did was to evaluate this limit:
    [tex] \lim_{k \to \infty} \frac {a_{k+1}}{a_k} = \lim_{k \to \infty} \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}= \lim_{k \to \infty} r \cdot \frac{k^r}{(k+1)^r } = \lim_{k \to \infty} r \cdot (\frac{k}{(k+1)})^r[/tex]
    What do you get for this limit? And what does that value imply for the test you are using?
     
  12. Mar 24, 2010 #11
    I get [tex]r \cdot 1^r = r[/tex]?
    And r is positive
     
  13. Mar 24, 2010 #12

    Mark44

    Staff: Mentor

    From your work, what are the values of r that cause this series to converge?
     
  14. Mar 24, 2010 #13
    I need the ratio to approach something < 1?
    So since r is positive I need it to be 0 < r < 1?
     
  15. Mar 24, 2010 #14

    Mark44

    Staff: Mentor

    C'mon, show some confidence. Tell me, don't ask me.
     
  16. Mar 24, 2010 #15

    Mark44

    Staff: Mentor

    C'mon, show a little confidence in your ability. Tell me, don't ask me.
     
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