Can you take the derivative of this?

[matt grime]

Which is false. Therefore' date=' your k wasn't arbitrary.

 

I'm not sure yet how that affects your argument, but your claim that k can be any arbitrary positive integer is false.

 

That might not matter, I'll have to think about it a little more.

.[/quote']

 

If you cannot immediately see that since n!^{1/n? is greater than any positive integer greater than 2 for n sufficiently large still implies that it diverges then you really are missing something.

[matt grime]

Can I offer a quick summary of your posting style? Post 23. The first 21 lines are unnecessary, being almost entirely summed up in line 22.

 

The rest was a very verbose and unnecessary rewrite of a reasonable short proof. Why? That is rhetorical and does not require you to answer. It is supposed to make you think about wasting time posting unneccesary junk. For the sake of poeple on dail up can you not post huge long posts with loads of latex in them that are completely unnecessary?

[Johnny5]

Johnny' date=' when I said, in my argument that you think is wrong, when k=1, that

 

n! >k^{n-k}

 

for all n sufficiently large, why do you ignore the fact that n! *is* strictly greater than 1 for all n sufficiently large? n greater than 2 in this case.

 

I did not state it was true for all n; it isn't; it is true for all n greater than some number dependent on k, but that is immaterial.

 

Perhaps the > should have been a => sign, I honestly can't remember exactly what I intended to type but it really is not important. Once more you miss the point of the argument and focus on unnecessary aspects, just like you missed the point about absolute convergence of power series. It is not important that the power series for (1+x)^t may converge for some t when x is 1, but that it diverges for |x|>1[/quote']

 

I do not think your argument is wrong Matt.

 

Yes, true for n dependent on k.

 

It should have been greater than or equal to, because you started out by letting k denote an arbitrary positive integer, and if k=1 then there is a false statement later in the proof. That's why you should have written >, but this was just a minor logical point. The main point is, that your proof works, and is easy to follow. I think it is a fine proof Matt.

[Johnny5]

The second proof is much easier: since exp(k) has a power series valied for all k in the real numbers it follows that k^n/n!' date=' the n'th term must converge to zero, that is, assuming k is positive,

 

k^n/n! <1 for all n sufficiently large, ie k^n<n! for all n sufficiently large.

 

But that requires you to know taylor series, radii of convergence and d'alembert's ratio test.

 

My proof can be followed merely by common sense and could be dreamt up by anyone who is prepared to think a little.[/quote']

 

I haven't gone through your second proof yet, but your first proof worked.

[Johnny5]

If you cannot immediately see that since n!^{1/n} is greater than any positive integer greater than 2 for n sufficiently large still implies that it diverges then you really are missing something.

 

 

No i see that, it's in post 23.

 

As I said somewhere, its been over a decade since I took advanced calculus. I did follow your proof. The only thing I had a problem with, was when you wrote >, instead of >, but that is of course just a minor error on your part.

 

Regards

 

PS: I'm glad to see the material again, its been a long time. You did great. Thank you

[matt grime]

If you're going to get bogged down in unimportant minor details you are never going to get anywhere. Oh, and it is also not the done thing to correct a mathematicians errors when they are obvious and unimportant. This may seem strange, and indeed disturbing to some people. One certainly would never refer to writing > instead of => as an error, minor or otherwise. One would instead refer to the writer making a typo or some such euphemism, and the writer would readily acknowledge the mistake and thank the person for pointing it out. When you say things like "this bit's wrong, and i don't know how it affects your argument" you show yourself up since it clearly doesn't affect the conclusion at all. Mathematicians make mistakes, lots of them; they are only serious if they are mistakes of understanding.

[Johnny5]

they are only serious if they are mistakes of understanding.

 

Pretty much i agree with that. Even though no one's perfect, it's fun to try though.

 

Regards.

 

Also, i was just reading this:

 

Limit of a sequence

Main article: limit of a sequence

 

Consider the following sequence: 1.79, 1.799, 1.7999,... We could observe that the numbers are "approaching" the 1.8, the limit of the sequence.

 

Formally, suppose x1, x2, ... is a sequence of real numbers. We say that the real number L is the limit of this sequence and we write

 

 

if and only if

 

for every ε>0 there exists a natural number n0 which will depend on ε such that for all n>n0 we have |xn - L| < ε.

 

And also read that a sequence has a limit only if it's a Cauchy sequence.

 

I did a hundred or so proofs using that definition above, but its been quite awhile since i've done one. Actually, maybe far more than that, difficult to say. Might have been 500, but my point is, that it's been awhile.

[matt grime]

A sequence of real numbers converges if and only if it is Cauchy, yes, that is true, and easy to prove since R is a complete metric space.

[fuhrerkeebs]

I did a hundred or so proofs using that definition above, but its been quite awhile since i've done one. Actually, maybe far more than that, difficult to say. Might have been 500, but my point is, that it's been awhile.

 

Ha.