Show that for any

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- October 2nd 2013, 05:55 PMvidomagruProve x > ln(1+x) for all x > 0
Show that for any

- October 2nd 2013, 06:19 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
Let . Now, you need to show for all . What is ? Knowing that, how would you show for all ?

- October 2nd 2013, 06:25 PMPlatoRe: Prove x > ln(1+x) for all x > 0
- October 2nd 2013, 06:41 PMvidomagruRe: Prove x > ln(1+x) for all x > 0
- October 2nd 2013, 06:59 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
You know that , so if for then the function must increase. Show that for . After that, you know that is continuous on and differentiable on for any . Use the Mean Value Theorem.

- October 2nd 2013, 07:18 PMvidomagruRe: Prove x > ln(1+x) for all x > 0
To show this let and finding the derivative of we have: which is clearly positive for all We know that and that is continuous on and differentiable on for any . So using the mean value theorem we have:

**I am not sure how to manipulate it from there. Is that right so far?** - October 2nd 2013, 07:45 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
- October 2nd 2013, 08:30 PMvidomagruRe: Prove x > ln(1+x) for all x > 0
So here is a revamped attempt at the whole underlying question:

__Problem:__

Let where

Prove that the sequence converges.

Hint: Show, first, that for any . Then show that the sequence decreases and . Use the theorem about the convergence and divergence of p-series to complete the proof.

*So here is what I have following the hint from above, but I have some blanks that I don't know how to fill in, and I am not sure if this is a good proof?*

First we show that .

To show this let and finding the derivative of we have: which is clearly positive for all We know that and that is continuous on and differentiable on for any . So using the mean value theorem we have:

Next, we know that there exists such that so We have shown that And since the product of two positive numbers is positive, is positive. This is true so we have shown that

Next we show that decreases and

We can show that decreases with induction. Claim: Consider Now consider Hence it holds for that Now suppose that is true. This means Now and since we have shown that , it follows that and therefore Therefore we have shown by the induction hypothesis that is decreasing over all

Next I am supposed to show that but I am not sure how or why this is important.

Help? Should I just be trying to use the monotone convergence theorem? - October 2nd 2013, 08:50 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
The last equality is not correct:

You only need the first equality:

Next, if decreases, then . I think you want to show .

Then, . You want to show . So, . I don't know how that shows you that your function is decreasing.

Alternately, if we look at , we have . So, you want to show that . For that, I am not sure how is helpful...

Edit: Oh, but .

So, you want to show . - October 3rd 2013, 04:52 AMSlipEternalRe: Prove x > ln(1+x) for all x > 0
Suppose . Then .

You already showed that . Next, let's show that that using the same method. Let . We have and , so for all . And again, using the Mean Value Theorem on any interval , we find that for some , so .

Hence for all , . So . This means

You don't quite get to the same place as the hint, but you do get that the different is less than . So, . Let's take the limit: . Let's look at the limit of the sum:

- October 3rd 2013, 07:35 AMHartlwRe: Prove x > ln(1+x) for all x > 0
x>in(1+x)

e^x>1+x

e^x = 1+x+.... >1+x - October 3rd 2013, 08:13 AMHartlwRe: Prove x > ln(1+x) for all x > 0
Please note latest edit of last post. I should have done a new post.

- October 3rd 2013, 09:03 AMSlipEternalRe: Prove x > ln(1+x) for all x > 0
I think the hint is wrong... We want to show that . (This is the original expression for ). Let's write out the power series:

So we have,

Edit: But, using this, it is easy to see that

I am not sure if that is any easier, though. - October 3rd 2013, 09:14 AMHartlwRe: Prove x > ln(1+x) for all x > 0
Just out of curiousity, what's wrong with:

x>in(1+x)

e^x>1+x

e^x = 1+x+.... >1+x - October 3rd 2013, 09:42 AMSlipEternalRe: Prove x > ln(1+x) for all x > 0
That depends on what you mean. When the order of those three statements is reversed, it is a valid proof, so long as you have already proven that implies (which is true).

Edit: The proof is extremely straightforward.

Let . Assume .

We know that . Consider the function . For all . So, the natural logarithm is increasing on the entire range of . Hence, implies .