Results 1 to 4 of 4

Math Help - nice inequality

  1. #1
    Junior Member
    Joined
    Jul 2010
    Posts
    25

    nice inequality

     <br /> <br />
\begin{array}{l}<br />
 prove\;that \\ <br />
  \\ <br />
 1 \prec \frac{1}{{1001}} + \frac{1}{{1002}} + \frac{1}{{1003}} + ...... + \frac{1}{{3009}} \prec 1\frac{1}{3} \\ <br />
 \end{array}<br /> <br /> <br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by fxs12 View Post
     <br /> <br />
\begin{array}{l}<br />
 prove\;that \\ <br />
  \\ <br />
 1 \prec \frac{1}{{1001}} + \frac{1}{{1002}} + \frac{1}{{1003}} + ...... + \frac{1}{{3009}} \prec 1\frac{1}{3} \\ <br />
 \end{array}<br /> <br /> <br />
    Euler's constant is

    \displaystyle\lim_{n\rightarrow\infty}\left(\frac{  1}{1}+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{n}-log_en\right)

    log_en\approx\left(1+\frac{1}{2}+\frac{1}{3}+....+  \frac{1}{n}\right)-Euler's\ constant


    \frac{1}{1001}+\frac{1}{1002}+.......+\frac{1}{300  9}\approx\ log_e3009-log_e1000\approx\ log_e3.009



    e=2.718...

    e^{\frac{4}{3}}=3.79...


    Therefore

    log_e2.718..<log_e3.009<log_e3.79..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2010
    Posts
    25
    thank you , but how didi you get this?


    \log _e n \approx (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ....... + \frac{1}{n}) - Euler's\;cons\tan t
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Sorry, that approximation may be a bit on the crude side!!

    What you have is the difference between a pair of Harmonic series.

    There are alternative quick calculations for that,
    and you may be requiring a particular type of solution,
    possibly along other lines.

    H_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..........  ......+\frac{1}{n}

    One approximation is...

    \frac{1}{2(n+1)}+log_en+\gamma<H_n<\frac{1}{2n}+lo  g_en+\gamma

    where \gamma\approx\ 0.5772156649 is Euler's constant.

    Taking either \frac{1}{2n} or \frac{1}{2n+1}

    we get approximately H_{3009}-H_{1000}\approx\ \frac{1}{2(3009)}+log_e3009-\left(\frac{1}{2(1000)}+log_e1000\right)

    or

    \frac{1}{2(3010)}+log_e3009-\left(\frac{1}{2(1001)}+log_e1000\right)

    which is basically approximately the difference between the logarithms
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Nice Inequality !!
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: February 6th 2010, 04:08 AM
  2. Nice limit
    Posted in the Calculus Forum
    Replies: 12
    Last Post: January 1st 2010, 12:40 PM
  3. Very nice inequality
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: June 20th 2008, 04:56 PM
  4. Replies: 21
    Last Post: November 11th 2007, 08:20 AM
  5. NIce one, who will do it first??
    Posted in the Algebra Forum
    Replies: 7
    Last Post: September 6th 2007, 07:48 AM

Search Tags


/mathhelpforum @mathhelpforum