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An infinite geometric series

  1. Apr 12, 2003 #1
    Can you solve analytically

    [pi] (n)1/n


    [pi] (n!)1/n!


    [sum] (1/n)n


    [sum] (1/n!)n!

  2. jcsd
  3. Apr 12, 2003 #2


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    &Pi n1/n = [oo]
    n = 1


    Let P_k be the k-th partial product of &Pi n1/n


    ln P_k = ln &Pi n1/n = &Sigma (1/n) ln n

    Notice that the n-th term in this sequence is strictly greater than 1/n (for all n > 1)... so the k-th partial sum must be greater than the k-th partial sum of the harmonic series. Since the harmonic series diverges, so must this.

  4. Apr 12, 2003 #3
    [sum](n!)1/n! = ∞

     ∞               ∞
    [sum](n!)1/n! > [sum] 1
    n=1            n=1
    The right hand side obviously diverges, since the left hand side is larger then then the right hand side it diverges also.  ♦

    ps: I edited this about 30 times to make it look nice and pretty, so you better all appreciate it. That line with the inequality was wicked hard to put together and make everything line up just right
    Last edited by a moderator: Apr 12, 2003
  5. Apr 13, 2003 #4
    Neat, Hurkyl. Although straightforward, I hadn't thought of taking the logarithm of the partial product to get a summation.

    What do you mathematicians think of climbhi's "proof"?

    Who can (re)evaluate the other infinite product and summations?
  6. Apr 13, 2003 #5
    Well admitidley I am not a mathemitician (or a great speller) but I think this is entirely valid. Just looking at the first few terms
    • a1 = 1 which is ≥ 1
    • a2 = √2 again ≥ 1
    • a3 = 61/6 again ≥ 1
    • a4 = 241/24 again ≥ 1
    • ....
    • an = n!1/n! since the limit of n!1/n! as n→∞ = 1 we can again say an ≥ 1

    So it's obvious that our sum of n!1/n! from 1 to ∞ is greater then our sum of 1 from 1 to ∞. Now since 1 ≥ 1/n for all natural n we know the sum of 1 from 1 to ∞ is greater then the harmonic series, which diverges, so so does the sum of ones, so so does the sum of n!1/n!. I really think that this is valid and have no idea why you're questioning it. Please tell me, have I done something wrong here?
  7. Apr 13, 2003 #6
    Well, lim sum (1/n)^n is somewhere near 1.29128599706266...
    This number looks familiar...
  8. Apr 13, 2003 #7


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    Climbhi's proof is solid, it's just not one of the 4 you asked for, hehe.

  9. Apr 13, 2003 #8
    I guess that lim sum (1/n)^n can't be written as a function of other constants, so it is a new constant...like lim 1+1/2+1/3+...+1/n-ln(n), which is equal to c in (0,1), Euler's constant...
    The same for lim sum (1/n!)^n!...

    lim sum (1/n)^n=L, 1.29<L<1.292; L ~ 1.29128599706266

    lim sum (1/n!)^n!=B, 0.29<B<0.292; B ~ 0.291285997062663

    L from Loren, and B from Booda...

    Even though I've said these things...that 1.29128599706266 looks very, very familiar to me...it could be a physical constant...

    I think you know how to prove that sum (1/n)^n is convergent...and sum (1/n!)^n!...
    If not...here it is...
    1 < sum (1/n)^n < sum (1/2)^(n-1) < 1.5;
    Let Ak = sum (1/n)^n...sum from 1 to k...
    Ak < A(k+1)...evidently...so...according to Weierstrass theorem (Ak is bounded (1,1.5) and it is raising) we obtain that the sum is convergent...the same for sum (1/n!)^n!...
  10. Apr 13, 2003 #9
    Oooh, Loren was looking for the infinite product, as you can see in my replies I thought it was the infinite sum of n!1/n! that he was looking for. Sorry about that!
    Hurkyl, would you mind explaining you're proof to me, it looks like in the middle you suddenly jumped from an infinite product to an infinite sum. Did you do this using the properties of logarithms (i.e. ln6 = ln3 + ln2) or am I missing something. Thanks.
  11. Apr 13, 2003 #10
    ln x^(1/x)=(1/x)*lnx
    and ln (1^1 * 2^(1/2) * ... * k^(1/k) ) is equal to
    ln 1 + (1/2)*ln2 + ... + (1/k)*lnk...
    I guess the harmonic series means 1/1+1/2+1/3+...+1/n...?
    (which is divergent...)
  12. Apr 13, 2003 #11
    Thank you, bogdan, my appreciation for you and your numerical skills is without limit. However, with 100,000 new math proofs every year...are these numbers original? "Loren" and "Booda" seem to represent properties similar to the Golden Means', perhaps here a polynomial with irrational coefficients. Say L=B+1, where x=L and x=B so x^2+(2B+1)x+B(B+1)=0.
  13. Apr 14, 2003 #12
    x^2-(2B+1)x+B(B+1)=0 ...
    Due to a programming error... B ~ 1.25002143347051...
    Last edited: Apr 14, 2003
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