- #1

Loren Booda

- 3,119

- 4

[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

?

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- Thread starter Loren Booda
- Start date

- #1

Loren Booda

- 3,119

- 4

[oo]

n=1

or

[oo]

n=1

or

[oo]

n=1

or

[oo]

n=1

?

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

&Pi n

n = 1

Proof:

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

Then:

ln P_k = ln &Pi 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

- #3

_{∞}

[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

[sum](n!)

Proof:

[sum](n!)

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:

- #4

Loren Booda

- 3,119

- 4

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

- #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 termsOriginally posted by Loren Booda

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

- a
_{1}= 1 which is ≥ 1 - a
_{2}= √2 again ≥ 1 - a
_{3}= 6^{1/6}again ≥ 1 - a
_{4}= 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!

- #6

bogdan

- 191

- 0

Well, lim sum (1/n)^n is somewhere near 1.29128599706266...

This number looks familiar...

This number looks familiar...

- #7

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 19

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

Hurkyl

Hurkyl

- #8

bogdan

- 191

- 0

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!...

- #9

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!

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.

- #10

bogdan

- 191

- 0

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...)

- #11

Loren Booda

- 3,119

- 4

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.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...

- #12

bogdan

- 191

- 0

x^2-(2B+1)x+B(B+1)=0 ...

Due to a programming error... B ~ 1.25002143347051...

Sorry...

Due to a programming error... B ~ 1.25002143347051...

Sorry...

Last edited:

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