How do you prove that ? Starting from the fact that , so without using anything related to .

And also that .

I just don't understand what my teacher did... Meaningless to me.

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- Jul 10th 2008, 10:01 AMarbolis[SOLVED] limit of the natural logarithm
How do you prove that ? Starting from the fact that , so without using anything related to .

And also that .

I just don't understand what my teacher did... Meaningless to me. - Jul 10th 2008, 10:35 AMTheEmptySet
- Jul 10th 2008, 10:45 AMarbolis
Thanks! I understand what you did.

- Jul 10th 2008, 11:10 AMinvite_moon
to see this

- Jul 10th 2008, 11:19 AMarbolis
Thanks invite moon! My professor did something similar to this, now I understand him.

In your explanation, you should also show that is increasing, and that . - Jul 10th 2008, 11:33 AMinvite_moon
Sorry ,I thought that ln(x) is increasing and that ln3>1,are well known things.

- Jul 10th 2008, 11:56 AMarbolis
Yes there are. For instance, , it can be shown that . The derivative of is , always positive in the dominion of the ln function, so ln is increasing.

- Jul 10th 2008, 01:03 PMPaulRS
First note that natural logarithm is an incresing function (1) since for all

Now:

Thus the Natual Logarithm has the following property, for :

Thus: for any natural number (2)

Suppose we want to prove that by definition.

We have to prove that, for all there is a number such that if we have (3)

Setting by (1) and (2) we know that the condition (3) holds.

Just see that iff and since the function is increasing if then

So by definition it follows that (4)

For the second part set and use the property shown in the second line, so that and since we have and the assertion follows by (4) - Jul 10th 2008, 01:31 PMarbolis
Nicely done PaulRS. Thank you.

- Jul 10th 2008, 02:42 PMThePerfectHacker
The problem with this approach is that the typical proof of the divergence of the harmonic series uses the integral test which uses . So it is kind of circular. But not really because there are some proofs which show divergence of harmonic series without integral tests.

Good job. But there is just a minor problem here. What does mean? You are assuming the definition of exponents for arbitrary real powers. However, the generalized definition needs further properties of which you are trying to avoid in the first place. - Jul 10th 2008, 06:27 PMNonCommAlg
- Jul 10th 2008, 07:51 PMThePerfectHacker
Here is a general theorem. Let be a function (defined on ) such that (i) is increasing (ii) is not bounded from above. Then . Here is the proof. We prove this directly by definition. Let . Then since is not bounded from above there is such that . Then if it means . Proof complete. Now we can apply this to function. Since this function is increasing this satisfies (i). Now we need to satisfy (ii) i.e. the function is unbounded this takes a little bit more work. Notice that (when ) and . Thus, . Since and it means there is a number, call it , such that (by property of continous functions). Then we see as

**PaulRS**showed that for all positive integers . But since we can make are large as we please it means is an unbounded function. The rest follows.