How to Solve for the Value of B in a Geometric Series

[Loren Booda]

Can you solve analytically

[oo]
_[pi] (n)^1/n
n=1

or

[oo]
_[pi] (n!)^1/n!
n=1

or

[oo]
_[sum] (1/n)ⁿ
n=1

or

[oo]
_[sum] (1/n!)^n!
n=1


?

 

[Hurkyl]

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

Proof:

Let P_k be the k-th partial product of &Pi n^1/n

Then:

ln P_k = ln &Pi n^1/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.


Hurkyl

 

[climbhi]

 _∞
[sum](n!)^1/n! = ∞
ⁿ⁼¹

Proof:
 _∞               _∞
[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

 

[Loren Booda]

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?

 

[climbhi]

Originally posted by Loren Booda
What do you mathematicians think of climbhi's "proof"?

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 = 6^1/6 again ≥ 1
• a₄ = 24^1/24 again ≥ 1
• ...
• a_n = n!^1/n! since the limit of n!^1/n! as n→∞ = 1 we can again say a_n ≥ 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?

 

[bogdan]

Well, lim sum (1/n)^n is somewhere near 1.29128599706266...
This number looks familiar...

 

[Hurkyl]

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

Hurkyl

 

[bogdan]

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!...

 

[climbhi]

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

Hurkyl

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.

 

[bogdan]

Yes...
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...)

 

[Loren Booda]

bogdan

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...

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.

 

[bogdan]

x^2-(2B+1)x+B(B+1)=0 ...
Due to a programming error... B ~ 1.25002143347051...
Sorry...