Show that $\displaystyle x > \ln(1+x)$ for any $\displaystyle x > 0$

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- Oct 2nd 2013, 04:55 PMvidomagruProve x > ln(1+x) for all x > 0
Show that $\displaystyle x > \ln(1+x)$ for any $\displaystyle x > 0$

- Oct 2nd 2013, 05:19 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
Let $\displaystyle f(x) = x - \ln(1+x)$. Now, you need to show $\displaystyle f(x)>0$ for all $\displaystyle x>0$. What is $\displaystyle f(0)$? Knowing that, how would you show $\displaystyle f(x)>0$ for all $\displaystyle x>0$?

- Oct 2nd 2013, 05:25 PMPlatoRe: Prove x > ln(1+x) for all x > 0
Do you know Napier's inequity?

If $\displaystyle 0<a<b$ then $\displaystyle \frac{1}{b} < \frac{{\log (b) - \log (a)}}{{b - a}} < \frac{1}{a}$. Easily proven by the mean value theorem.

Into that inequality substitute $\displaystyle b=x+1~\&~a=1$. What you ask falls right out. - Oct 2nd 2013, 05:41 PMvidomagruRe: Prove x > ln(1+x) for all x > 0
- Oct 2nd 2013, 05:59 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
You know that $\displaystyle f(0)=0$, so if $\displaystyle f(x)>0$ for $\displaystyle x>0$ then the function must increase. Show that $\displaystyle f'(x)>0$ for $\displaystyle x>0$. After that, you know that $\displaystyle f(x)$ is continuous on $\displaystyle [0,a]$ and differentiable on $\displaystyle (0,a)$ for any $\displaystyle a>0$. Use the Mean Value Theorem.

- Oct 2nd 2013, 06:18 PMvidomagruRe: Prove x > ln(1+x) for all x > 0
To show this let $\displaystyle f(x) = x - \ln(1+x);$ and finding the derivative of $\displaystyle f(x)$ we have: $\displaystyle f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x}$ which is clearly positive for all $\displaystyle x > 0.$ We know that $\displaystyle f(0) = 0$ and that $\displaystyle f(x)$ is continuous on $\displaystyle [0,a]$ and differentiable on $\displaystyle (0,a)$ for any $\displaystyle a>0$. So using the mean value theorem we have:

$\displaystyle \frac{f(a) - f(0)}{a-0} = \frac{f(a)}{a} = \frac{a - \ln(1+a)}{a} = \frac{x}{1+x}.$

**I am not sure how to manipulate it from there. Is that right so far?** - Oct 2nd 2013, 06:45 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
Yes, that is right so far. Next, you know there exists $\displaystyle c\in (0,a)$ such that $\displaystyle f'(c) = \frac{f(a)}{a}$ so $\displaystyle f(a) = a\cdot f'(c)$. You just showed that $\displaystyle f'(c)>0$ and this is true for any $\displaystyle a>0$. The product of two positive numbers is positive. So, $\displaystyle f(a)$ is positive. Again, this is true for any $\displaystyle a>0$. So, you are done.

- Oct 2nd 2013, 07:30 PMvidomagruRe: Prove x > ln(1+x) for all x > 0
So here is a revamped attempt at the whole underlying question:

__Problem:__

Let $\displaystyle x_n = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} - \ln(n)$ where $\displaystyle n = 1,2,3,...$

Prove that the sequence $\displaystyle {x_n}$ converges.

Hint: Show, first, that $\displaystyle x > \ln(1+x)$ for any $\displaystyle x > 0$. Then show that the sequence $\displaystyle {x_n}$ decreases and $\displaystyle x_{n+1} - x_n < \frac{1}{2(n+1)^2}$. 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 $\displaystyle x > \ln(1+x)$.

To show this let $\displaystyle f(x) = x - \ln(1+x);$ and finding the derivative of $\displaystyle f(x)$ we have: $\displaystyle f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x}$ which is clearly positive for all $\displaystyle x > 0.$ We know that $\displaystyle f(0) = 0$ and that $\displaystyle f(x)$ is continuous on $\displaystyle [0,a]$ and differentiable on $\displaystyle (0,a)$ for any $\displaystyle a>0$. So using the mean value theorem we have:

$\displaystyle \frac{f(a) - f(0)}{a-0} = \frac{f(a)}{a} = \frac{a - \ln(1+a)}{a} = \frac{x}{1+x}.$

Next, we know that there exists $\displaystyle c \in (0,a)$ such that $\displaystyle f'(c) = \frac{f(a)}{a}$ so $\displaystyle f(a) = a \dot f'(c).$ We have shown that $\displaystyle f'(c) > 0, \forall a > 0.$ And since the product of two positive numbers is positive, $\displaystyle f(a)$ is positive. This is true $\displaystyle \forall a > 0$ so we have shown that $\displaystyle x > \ln(1+x).$

Next we show that $\displaystyle {x_n}$ decreases and $\displaystyle x_{n+1} - x_n < \frac{1}{2(n+1)^2}.$

We can show that $\displaystyle {x_n}$ decreases with induction. Claim: $\displaystyle x_n \ge x_{n+1}.$ Consider $\displaystyle n=1: x_1 = 1 - \ln(1) = 1 - 0 = 1.$ Now consider $\displaystyle x_{n+1} = x_{1+1} = x_2 = 1 + \frac{1}{2} - \ln(2) = 0.802.$ Hence it holds for $\displaystyle n=1$ that $\displaystyle x_n \ge x_{n+1}.$ Now suppose that $\displaystyle n=k$ is true. This means $\displaystyle x_k \ge x_{k+1}.$ Now $\displaystyle x_{(k+1)+1} = 1 + \frac{1}{2} + ... + \frac{1}{k} + \frac{1}{k+1} - \ln(k+1) = x_{k+1} + \frac{1}{k+1} - \ln(k+1)$ and since we have shown that $\displaystyle x > \ln(1+x)$, it follows that $\displaystyle x_{k+1} + \frac{1}{k+1} - \ln(k+1) \ge x_{k+1}$ and therefore $\displaystyle x_{(k+1)+1} \ge x_{k+1}.$ Therefore we have shown by the induction hypothesis that $\displaystyle {x_n}$ is decreasing over all $\displaystyle n \in \mathbb{N}.$

Next I am supposed to show that $\displaystyle x_{n+1} - x_n < \frac{1}{2)(n+1)^2}$ but I am not sure how or why this is important.

Help? Should I just be trying to use the monotone convergence theorem? - Oct 2nd 2013, 07:50 PMSlipEternalRe: Prove x > ln(1+x) for all x > 0
The last equality is not correct: $\displaystyle \frac{f(a) - f(0)}{a-0} = \frac{f(a)}{a} = \frac{a - \ln(1+a)}{a} = \frac{x}{1+x}.$

You only need the first equality: $\displaystyle \frac{f(a) - f(0)}{a-0} = \frac{f(a)}{a}$

Next, if $\displaystyle x_n$ decreases, then $\displaystyle x_{n+1}-x_n < 0 < \dfrac{1}{2(n+1)^2}$. I think you want to show $\displaystyle |x_{n+1}-x_n| = x_n-x_{n+1} < \dfrac{1}{2(n+1)^2}$.

Then, $\displaystyle \begin{align*}x_{(k+1)+1} & = 1 + \dfrac{1}{2} + \ldots + \dfrac{1}{k+1} + \dfrac{1}{k+2} - \ln(k+2) \\ & = x_{k+1} + \dfrac{1}{k+2} + \ln(k+1) - \ln(k+2)\end{align*}$. You want to show $\displaystyle x_{k+2} < x_{k+1}$. So, $\displaystyle x_{k+2} = x_{k+1} + \dfrac{1}{k+2} + \ln \left( \dfrac{k+1}{k+2} \right) < x_{k+1} + \dfrac{1}{k+2} $. I don't know how that shows you that your function is decreasing.

Alternately, if we look at $\displaystyle x_{n+1}-x_n$, we have $\displaystyle x_{n+1} - x_n = x_n + \dfrac{1}{n+1} + \ln\left( \dfrac{n}{n+1} \right) - x_n = \dfrac{1}{n+1} + \ln\left( \dfrac{n}{n+1} \right)$. So, you want to show that $\displaystyle \left| \dfrac{1}{n+1} + \ln\left( \dfrac{n}{n+1} \right) \right| < \dfrac{1}{2(n+1)^2}$. For that, I am not sure how $\displaystyle x>\ln(x+1)$ is helpful...

Edit: Oh, but $\displaystyle \dfrac{1}{n+1} + \ln\left( \dfrac{n}{n+1} \right) = \dfrac{1}{n+1} - \ln\left( 1 + \dfrac{1}{n} \right)$.

So, you want to show $\displaystyle \ln\left( 1 + \dfrac{1}{n} \right) < \dfrac{1}{2(n+1)^2} + \dfrac{1}{n+1}$. - Oct 3rd 2013, 03:52 AMSlipEternalRe: Prove x > ln(1+x) for all x > 0
Suppose $\displaystyle x = \dfrac{1}{n}$. Then $\displaystyle \dfrac{x}{1+x} = \dfrac{\tfrac{1}{n}}{1 + \tfrac{1}{n}} = \dfrac{1}{n+1}$.

You already showed that $\displaystyle x>\ln(1+x)$. Next, let's show that that $\displaystyle \ln(1+x)>\dfrac{x}{1+x}$ using the same method. Let $\displaystyle g(x) = \ln(1+x) - \dfrac{x}{1+x}$. We have $\displaystyle g(0) = 0$ and $\displaystyle g'(x) = \dfrac{1}{1+x} - \dfrac{1}{(1+x)^2} = \dfrac{x}{(1+x)^2}$, so $\displaystyle g'(x)>0$ for all $\displaystyle x>0$. And again, using the Mean Value Theorem on any interval $\displaystyle [0,a]$, we find that $\displaystyle \dfrac{g(a)-g(0)}{a-0} = \dfrac{g(a)}{a} = f'(c)$ for some $\displaystyle c \in (0,a)$, so $\displaystyle g(a) = a f'(c) > 0$.

Hence for all $\displaystyle x>0$, $\displaystyle \dfrac{x}{1+x} < \ln(1+x) < x$. So $\displaystyle \dfrac{1}{1+n} - \ln \left( 1 + \dfrac{1}{n} \right) < 0$. This means

$\displaystyle \begin{align*}|x_{n+1}-x_n| & = \left|\dfrac{1}{1+n} - \ln \left( 1 + \dfrac{1}{n} \right) \right| \\ & = \ln \left( 1 + \dfrac{1}{n} \right) - \dfrac{1}{1+n} \\ & < \dfrac{1}{n} - \dfrac{1}{1+n} \\ & = \dfrac{1}{n(1+n)} \\ & < \dfrac{1}{n^2}\end{align*}$

You don't quite get to the same place as the hint, but you do get that the different is less than $\displaystyle \dfrac{1}{n^2}$. So, $\displaystyle x_n = x_1 - \sum_{k=1}^{n-1}\left[\ln\left(1 + \dfrac{1}{k}\right) - \dfrac{1}{k+1} \right]$. Let's take the limit: $\displaystyle \lim_{n\to \infty} x_n = x_1 - \lim_{n \to \infty} \sum_{k=1}^{n-1}\left[\ln\left(1 + \dfrac{1}{k}\right) - \dfrac{1}{k+1} \right]$. Let's look at the limit of the sum:

$\displaystyle \begin{align*}\lim_{n \to \infty}\sum_{k=1}^{n-1}0 & \le \lim_{n \to \infty} \sum_{k=1}^{n-1}\left[\ln\left(1 + \dfrac{1}{k}\right) - \dfrac{1}{k+1} \right] \le & \lim_{n \to \infty} \sum_{k=1}^{n-1}\dfrac{1}{k^2} \\ 0 & \le \lim_{n \to \infty} \sum_{k=1}^{n-1}\left[\ln\left(1 + \dfrac{1}{k}\right) - \dfrac{1}{k+1} \right] \le & \dfrac{\pi^2}{6}\end{align*}$ - Oct 3rd 2013, 06:35 AMHartlwRe: Prove x > ln(1+x) for all x > 0
x>in(1+x)

e^x>1+x

e^x = 1+x+.... >1+x - Oct 3rd 2013, 07:13 AMHartlwRe: Prove x > ln(1+x) for all x > 0
Please note latest edit of last post. I should have done a new post.

- Oct 3rd 2013, 08:03 AMSlipEternalRe: Prove x > ln(1+x) for all x > 0
I think the hint is wrong... We want to show that $\displaystyle \left| \dfrac{1}{1+n} + \ln\left( \dfrac{n}{1+n} \right) \right| < \dfrac{1}{2(1+n)^2}$. (This is the original expression for $\displaystyle x_{n+1}-x_n$). Let's write out the power series:

$\displaystyle \begin{align*}\ln\left( \dfrac{n}{1+n} \right) & = \ln\left( 1 - \dfrac{1}{1+n} \right) \\ & = \sum_{k \ge 1} (-1)^{k+1}(-1)^k \dfrac{1}{k(n+1)^k} \\ & = - \sum_{k\ge 1} \dfrac{1}{k(n+1)^k} \\ & = -\dfrac{1}{n+1} - \dfrac{1}{2(n+1)^2} - \dfrac{1}{3(n+1)^3} - \cdots\end{align*}$

So we have,

$\displaystyle \begin{align*}\left|\dfrac{1}{1+n} - \dfrac{1}{1+n} - \dfrac{1}{2(n+1)^2} - \dfrac{1}{3(n+1)^3} - \cdots \right| & = \dfrac{1}{2(n+1)^2} + \dfrac{1}{3(n+1)^3} + \cdots \\ & > \dfrac{1}{2(n+1)^2} \end{align*}$

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

$\displaystyle \begin{align*}\left| x_{n+1} - x_n\right| & = \sum_{k\ge 2} \dfrac{1}{k(1+n)^k} \\ & = \dfrac{1}{1+n} \sum_{k\ge 1} \dfrac{1}{(k+1)(1+n)^k} \\ & < \dfrac{1}{1+n} \sum_{k\ge 1} \dfrac{1}{k(1+n)^k} \\ & = -\dfrac{1}{1+n}\ln\left(1 - \dfrac{1}{1+n}\right) \\ & = \dfrac{1}{1+n}\ln\left(1 + \dfrac{1}{n}\right)\end{align*}$

I am not sure if that is any easier, though. - Oct 3rd 2013, 08:14 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 - Oct 3rd 2013, 08:42 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 $\displaystyle e^x > e^y$ implies $\displaystyle x>y$ (which is true).

Edit: The proof is extremely straightforward.

Let $\displaystyle x,y\in \mathbb{R}$. Assume $\displaystyle e^x>e^y$.

We know that $\displaystyle \forall k \in \mathbb{R}, e^k>0$. Consider the function $\displaystyle \ln(k)$. For all $\displaystyle k>0, \dfrac{d}{dk}\ln(k) = \dfrac{1}{k}>0$. So, the natural logarithm is increasing on the entire range of $\displaystyle e^k$. Hence, $\displaystyle e^x>e^y$ implies $\displaystyle x = \ln(e^x) > \ln(e^y) = y$.