Page 1 of 2 12 LastLast
Results 1 to 15 of 26
Like Tree10Thanks

Math Help - Prove x > ln(1+x) for all x > 0

  1. #1
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Prove x > ln(1+x) for all x > 0

    Show that x > \ln(1+x) for any x > 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

    Let f(x) = x - \ln(1+x). Now, you need to show f(x)>0 for all x>0. What is f(0)? Knowing that, how would you show f(x)>0 for all x>0?
    Thanks from topsquark and Shakarri
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by vidomagru View Post
    Show that x > \ln(1+x) for any x > 0
    Do you know Napier's inequity?
    If 0<a<b then \frac{1}{b} < \frac{{\log (b) - \log (a)}}{{b - a}} < \frac{1}{a}. Easily proven by the mean value theorem.

    Into that inequality substitute b=x+1~\&~a=1. What you ask falls right out.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by SlipEternal View Post
    Let f(x) = x - \ln(1+x). Now, you need to show f(x)>0 for all x>0. What is f(0)? Knowing that, how would you show f(x)>0 for all x>0?
    Is this right?
    f(0) = 0 - \ln(1 + 0) = \ln(1) = 0. By definition of \ln(x), \ln(x+1) is always positive for all x > 0. So...?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

    You know that f(0)=0, so if f(x)>0 for x>0 then the function must increase. Show that f'(x)>0 for x>0. After that, you know that f(x) is continuous on [0,a] and differentiable on (0,a) for any a>0. Use the Mean Value Theorem.
    Thanks from vidomagru and topsquark
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by SlipEternal View Post
    You know that f(0)=0, so if f(x)>0 for x>0 then the function must increase. Show that f'(x)>0 for x>0. After that, you know that f(x) is continuous on [0,a] and differentiable on (0,a) for any a>0. Use the Mean Value Theorem.
    To show this let f(x) = x - \ln(1+x); and finding the derivative of f(x) we have: f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} which is clearly positive for all x > 0. We know that f(0) = 0 and that f(x) is continuous on [0,a] and differentiable on (0,a) for any a>0. So using the mean value theorem we have:

    \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?
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

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

    \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?
    Yes, that is right so far. Next, you know there exists c\in (0,a) such that f'(c) = \frac{f(a)}{a} so f(a) = a\cdot f'(c). You just showed that f'(c)>0 and this is true for any a>0. The product of two positive numbers is positive. So, f(a) is positive. Again, this is true for any a>0. So, you are done.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2013
    From
    Washington state
    Posts
    77
    Thanks
    1

    Re: Prove x > ln(1+x) for all x > 0

    So here is a revamped attempt at the whole underlying question:

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

    Prove that the sequence {x_n} converges.

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

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

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

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

    Next I am supposed to show that 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?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by vidomagru View Post
    So here is a revamped attempt at the whole underlying question:

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

    Prove that the sequence {x_n} converges.

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

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

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

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

    Next I am supposed to show that 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?
    The last equality is not correct: \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: \frac{f(a) - f(0)}{a-0} = \frac{f(a)}{a}

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

    Then, \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 x_{k+2} < x_{k+1}. So, 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 x_{n+1}-x_n, we have 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 \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 x>\ln(x+1) is helpful...

    Edit: Oh, but \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 \ln\left( 1 + \dfrac{1}{n} \right) < \dfrac{1}{2(n+1)^2} + \dfrac{1}{n+1}.
    Last edited by SlipEternal; October 2nd 2013 at 08:17 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

    Suppose x = \dfrac{1}{n}. Then \dfrac{x}{1+x} = \dfrac{\tfrac{1}{n}}{1 + \tfrac{1}{n}} = \dfrac{1}{n+1}.

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

    Hence for all x>0, \dfrac{x}{1+x} < \ln(1+x) < x. So \dfrac{1}{1+n} - \ln \left( 1 + \dfrac{1}{n} \right) < 0. This means
    \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 \dfrac{1}{n^2}. So, 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: \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:

    \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*}
    Last edited by SlipEternal; October 3rd 2013 at 04:01 AM.
    Thanks from vidomagru
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    x>in(1+x)
    e^x>1+x
    e^x = 1+x+.... >1+x
    Last edited by Hartlw; October 3rd 2013 at 06:45 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    Please note latest edit of last post. I should have done a new post.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

    I think the hint is wrong... We want to show that \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 x_{n+1}-x_n). Let's write out the power series:

    \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,

    \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
    \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.
    Last edited by SlipEternal; October 3rd 2013 at 08:37 AM.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: 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
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,880
    Thanks
    742

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by Hartlw View Post
    Just out of curiousity, what's wrong with:

    x>in(1+x)
    e^x>1+x
    e^x = 1+x+.... >1+x
    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 e^x > e^y implies x>y (which is true).

    Edit: The proof is extremely straightforward.

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

    We know that \forall k \in \mathbb{R}, e^k>0. Consider the function \ln(k). For all k>0, \dfrac{d}{dk}\ln(k) = \dfrac{1}{k}>0. So, the natural logarithm is increasing on the entire range of e^k. Hence, e^x>e^y implies x = \ln(e^x) > \ln(e^y) = y.
    Last edited by SlipEternal; October 3rd 2013 at 09:09 AM.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Prove this ...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 10th 2009, 09:47 AM
  2. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  3. prove
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 24th 2008, 07:15 PM
  4. Prove
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 24th 2008, 07:12 AM
  5. Prove
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 22nd 2008, 09:32 AM

Search Tags


/mathhelpforum @mathhelpforum