1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving Inequality

  1. Feb 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove that for any naturam number n > 1 :
    [itex] \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} > \frac{13}{24}[/itex]

    2. Relevant equations
    Not sure

    3. The attempt at a solution
    [itex] \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} > \frac{1}{2n} + \frac{1}{2n} + \frac{1}{2n} + ... + \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2}[/itex]

    Then I stuck...

    Thanks
     
  2. jcsd
  3. Feb 2, 2017 #2
    I'm not sure what approach your instructor had in mind in setting the problem, but unless I'm mistaken, we can actually prove a stronger inequality bound.

    If we define
    [tex]f(n) = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n},[/tex]
    then it is relatively straightforward to show that ##f(n+1) > f(n) \,\, \forall \,\, n##, and so it must be the case that ##f(n) \geq f(2)##.

    It is also possible to prove that ##\lim_{n \to \infty} f(n) = \ln 2## as an interesting exercise.
     
  4. Feb 2, 2017 #3

    Mark44

    Staff: Mentor

    You can do this by mathematical induction, which might be the strategy your instructor had in mind.
    Proving a base case with n = 2 or n = 3 is straightforward.
    Then, assume that the proposition is true for n = k; i.e., that ##\frac 1 {k + 1} + \frac 1 {k + 2} + \dots + \frac 1 {2k} > \frac {13} {24}## (the induction hypothesis).
    Finally, use the induction hypothesis to show that ##\frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2} > \frac {13} {24}## must be true, as well.
     
  5. Feb 3, 2017 #4
    Let me try:
    (i) For n = 2
    1/3 + 1/4 = 7/12 = 14/24 > 13/24 so it is true for n = 2

    (ii) Assume it is true for n = k
    ##\frac 1 {k + 1} + \frac 1 {k + 2} + \dots + \frac 1 {2k} > \frac {13} {24}##

    (iii) For n = k + 1
    ##\frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2}##

    ## = \frac 1 {k + 1} + \frac 1 {k + 2} + \frac 1 {k + 3} + \dots + \frac 1 {2k} + \frac 1 {2k + 1} + \frac 1 {2k + 2} - \frac 1 {k + 1}##

    ## > \frac {13} {24} + \frac 1 {(2k + 1) (2k + 2)} > \frac {13} {24}##

    Is this correct?
     
  6. Feb 3, 2017 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks good.
     
  7. Feb 12, 2017 #6
    Sorry for late reply

    Thank you very much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proving Inequality
  1. Prove this inequality (Replies: 2)

  2. Prove this inequality (Replies: 1)

  3. Proving this inequality (Replies: 21)

  4. An inequality to prove (Replies: 16)

Loading...