# How big is this number?

## Main Question or Discussion Point

How "big" is this number?

I have had absolutely no sleep for a while, so my math brain has been failing me. Last night, I was working on a problem and I believe that one of my connections is wrong. Is this number countably infinite or uncountably infinite (as m approaches infinity):

$\sum_{n=1}^{m}{(\frac{m!}{n!})^{a}}$

I feel like it is countably infinite, but I am not sure.

Also, what about just simply (m!)a as m approaches infinity? If it is uncountable, that will be rather cool, too, as that is what one of my proofs hinge on.

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I have had absolutely no sleep for a while, so my math brain has been failing me. Last night, I was working on a problem and I believe that one of my connections is wrong. Is this number countably infinite or uncountably infinite (as m approaches infinity):

$\sum_{n=1}^{m}{(\frac{m!}{n!})^{a}}$

I feel like it is countably infinite, but I am not sure.

Also, what about just simply (m!)a as m approaches infinity? If it is uncountable, that will be rather cool, too, as that is what one of my proofs hinge on.
There seems to be a little confusion. Numbers aren't countable or uncountable. Any particular number is just a number. The adjectives "countable" and "uncountable" refer to sets of numbers. So the rational numbers are countable, and the irrational numbers are uncountable. What that means is that the entire set of rationals is a countable set; and the entire set of irrationals is an uncountable set.

Does that make sense?

Also, what is a? Is it an arbitrary real? An integer? Or what?

I have had absolutely no sleep for a while, so my math brain has been failing me. Last night, I was working on a problem and I believe that one of my connections is wrong. Is this number countably infinite or uncountably infinite (as m approaches infinity):

$\sum_{n=1}^{m}{(\frac{m!}{n!})^{a}}$

I feel like it is countably infinite, but I am not sure.

Also, what about just simply (m!)a as m approaches infinity? If it is uncountable, that will be rather cool, too, as that is what one of my proofs hinge on.
If "a" is positive the sum is infinite, but if "a" is negative that is another story, but I feel that is infinite also.

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mathman

If a < 0, the sum converges to 1.

The series is 1 + 1/m + {1/m(m-1) + 1/m(m-1)(m-2) + .... !/m!}
The part in brackets ~< 1/m, So the limit is simply 1.

If a < 0, the sum converges to 1.

The series is 1 + 1/m + {1/m(m-1) + 1/m(m-1)(m-2) + .... !/m!}
The part in brackets ~< 1/m, So the limit is simply 1.[/QUOTET]
Thanks

Sorry, I am still blaming my lack of sleep for me not being clear. What I mean is something more along these lines...

If I have a set that has a cardinality of (m!)a as m approaches infinity and where a is some natural number >1, is the cardinality countable or uncountable? If the set is uncountably infinite, then could I say that:

S1 has a cardinality of the sum given in the first post
S2 has a cardinality of (m!)a (with the limit as m approaches infinity)

Then would |S1|/|S2| be irrational? If so, then would this show that ζ(s) is irrational for s>1 and is a natural number?

I argue that it would be irrational since |S1| is outside the set of integers and so is |S2|, so their ratio, in terms of c/d would not be the ratio of two integers, and thus not rational.

Sorry, I am still blaming my lack of sleep for me not being clear. What I mean is something more along these lines...

If I have a set that has a cardinality of (m!)a as m approaches infinity and where a is some natural number >1, is the cardinality countable or uncountable? If the set is uncountably infinite, then could I say that:

S1 has a cardinality of the sum given in the first post
S2 has a cardinality of (m!)a (with the limit as m approaches infinity)

Then would |S1|/|S2| be irrational? If so, then would this show that ζ(s) is irrational for s>1 and is a natural number?

I argue that it would be irrational since |S1| is outside the set of integers and so is |S2|, so their ratio, in terms of c/d would not be the ratio of two integers, and thus not rational.
I dont think that infinity over infinity is either rational or irrational snce it is not a number.

haruspex
Homework Helper
Gold Member

If I have a set that has a cardinality of (m!)a as m approaches infinity and where a is some natural number >1, is the cardinality countable or uncountable?
If you have an infinite sequence of sets (a sequence necessarily being countably infinite), and each set is finite, then the union of the sets must be countable.

MathematicalPhysicist
Gold Member

If a < 0, the sum converges to 1.

The series is 1 + 1/m + {1/m(m-1) + 1/m(m-1)(m-2) + .... !/m!}
The part in brackets ~< 1/m, So the limit is simply 1.
I believe you have a mistake, it should be:

(m!)^a+(m!/2)^a+(m!/3!)^a+....+(m!/((m-1)!))^a+1

but as you said for a<0, all the terms vanish as m->\infty except 1.

I believe you have a mistake, it should be:

(m!)^a+(m!/2)^a+(m!/3!)^a+....+(m!/((m-1)!))^a+1

but as you said for a<0, all the terms vanish as m->\infty except 1.
Raised to the power of -1 and reading from right to left, it is the same as Mathman wrote.