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B summation of series

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  1. Feb 13, 2019 at 9:00 AM #1
    I've seen the proof that the sum of 1/n for = 1 to infinity is infinity (which still blows my mind a little).

    Is the sum of 1/nn for n = 1 to infinity also infinity?

    i.e, 1 + 2/4 + 3/27 + 4/256+...
     
  2. jcsd
  3. Feb 13, 2019 at 9:03 AM #2

    jedishrfu

    Staff: Mentor

    I don't think you mean ##n^n## where ##n=1## right?

    also your series should read ##1 + 1/4 + 1/27 ... ##?
     
  4. Feb 13, 2019 at 9:04 AM #3

    mfb

    Staff: Mentor

    After the first two elements it is smaller than the sum over ##\displaystyle \frac{1}{2^n}## where you can find the sum directly.

    If you want to explore the region between divergence and convergence more, have a look at ##\displaystyle \frac{1}{n^2}## and ##\displaystyle \frac{1}{n \log(n)}## and ##\displaystyle \frac{1}{n \log(n)^2}##and ##\displaystyle \frac{1}{n \log(n) \log(log(n))}##. These will need some clever tricks to determine if they converge.
     
  5. Feb 13, 2019 at 9:19 AM #4
    oops, yes, thanks, I meant the sum of n/nn for n = 1 to infinity

    So... it's < 2?
     
  6. Feb 13, 2019 at 9:29 AM #5

    mfb

    Staff: Mentor

    ##\frac{n}{n^n}##? That is equal to ##\frac{1}{n^{n-1}}##
    It is.
     
  7. Feb 13, 2019 at 1:27 PM #6

    BWV

    User Avatar

    the sum of ak converges for a<1 for k=1 to inf - this is just a geometric series and the sum is 1/(1-a)

    so by that 1/nn converges for n=1 to inf as 1/nn becomes smaller than any constant ak
     
  8. Feb 13, 2019 at 2:09 PM #7
    He meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n}[/itex].
     
  9. Feb 13, 2019 at 2:57 PM #8
    Same thing?
     
  10. Feb 14, 2019 at 1:22 PM #9
    Unless you guys are mind readers, and meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n} = \sum\limits_{n\to\infty} \,\, \frac{1}{n^{n-1}}[/itex] by what you said, [itex]\frac{n}{n^n}\, =\, \frac{1}{n^{n-1}}[/itex] is different from the sum.
     
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