Skip to main content

Inequality Problem


Source:
Posted by Shubham Agarwal (B.Tech CSE, NIT Raipur) on his blog

Problem:
Prove the following inequality:
   1/2 . 3/4 . 5/6 . .... 99/100 < 1/10

Bonus: Prove the generalized inequality: 
   1/2 . 3/4 . 5/6 . .... (2n-1)/2n < 1 / sqrt(2n)

Update: (12 Sep 2012)
3 different solutions posted in comments by sriram, chetan, dvdreddy, insomniac and sarat

Update: (4 Feb 2013)
Solution by Sriram and Chetan needs explanation. So, we have 2 different solutions by dvdreddy, insomniac and sarat.


Comments

  1. Straightforward proof, by induction!!! for all n>=2

    ReplyDelete
  2. Say P1=(3/2)(5/4)…(97/96)(99/98) // Has 49 fractions
    and P2=(2/1)(4/3)(6/5)…(98/97)(100/99) // Has 50 fractions

    Now you can see P1*P2= 100 and you can see that P2 > P1 and so P2 > sqrt(100) = 10
    and so 1/P2 < 1/10 which is the required result.

    You can see that it can be easily extended to the general inequality.

    Hope its correct.

    Thanks,
    Saratchand

    ReplyDelete
  3. Through induction... it can be assumed right till n-1 cases since the base case is true, then we need to prove 1/sqrt(2n-2) . 2n-1/2n<1 / sqrt(2n) implying a^2<(a-1)*(a+1) which is trivial

    ReplyDelete
    Replies
    1. how is a^2 < a^2 - 1? I think your final statement is false

      Delete
    2. I agree. Thanks for pointing that out Sriram. Can you please provide your induction argument please? Thanks

      Delete
  4. LHS = (2n Choose n)/4^n

    Stirling's formula will be used to simplify the LHS. In particular, [1] and [2] provide bounds in the form of a double inequality for n!

    n! < sqrt(2*pi) n^(n + 0.5) e^(-n + 1/12n)

    n! > sqrt(2*pi) n^(n + 0.5) e^(-n + 1/(12n + 1))

    These bounds are motivated from the Stirling's formula. Since we want to prove an inequality, we can't directly use Stirling's formula as it's only an approximation.

    Using these bounds, we have,

    LHS
    = (2n Choose n)/ 4^n
    < [1/sqrt(2*pi)](2n)^(2n + 0.5) e^(-2n + 1/24n) n^(-2n -1) e^(2n -2/(12n + 1))

    < [1/sqrt(pi*n)] e^(1/24n - 2/(12n+1))

    < [1/sqrt(pi*n)] for n>=1

    < [1/sqrt(2*n)] since pi > 2

    DONE.

    PEACE.


    References.

    [1] Robbins, H. "A Remark of Stirling's Formula." Amer. Math. Monthly 62, 26-29, 1955.


    [2] Feller, W. "Stirling's Formula." §2.9 in An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed. New York: Wiley, pp. 50-53, 1968.

    ReplyDelete
  5. Can be done using the sterling's Approximation
    which also gives the lower and upper bounds
    http://en.wikipedia.org/wiki/Stirling's_approximation

    1/2 . 3/4 . 5/6 . .... (2n-1)/2n
    can be written as below
    (2n)! / (pow(2, 2n) * n! * n1)

    now take log for the above expression

    we get ln(2n!) - 2n ln2 - 2 * ln(n!)

    now for 2n! use upper bound and n! use lower bound

    ReplyDelete
  6. We have 3 solutions to the problem :)

    Solution 1 - Induction: Thanks Sriram and Chetan
    Solution 2 - Stirling Approximation - Thanks dvdreddy and insomniac
    Solution 3 - Elementary Math - Thanks Sarat

    ReplyDelete
    Replies
    1. Correction. We do not have the induction solution by Sriram and Chetan

      Delete
  7. You could use AM-GM to prove the same.

    Let P be the product of 1/2, 3/4, ..., 2n-1/2n and S be their sum. AM-GM implies (S/n)^n > P.

    Now, S = sum (2i-1/2) = n - H(n)/2 where H(n): nth harmonic sum
    S/n = 1 - H(n)/2n < 1 - 1/2n since H(n) > 1

    P < (S/n)^n < (1 - 1/2n)^n < 1/sqrt (2n)

    ReplyDelete
    Replies
    1. Minor correction:
      S = sum (2i-1/2i) and not sum (2i-1/2)

      Please explain the statement (1 - 1/2n)^n < 1/sqrt (2n) in more detail. Thanks

      Delete
  8. a/b < a+1/b+1

    1/2.3/4.5/6.....99/100 < 2/3.4/5.6/7......100/101

    multiplying both sides by 1/2.3/4.5/6.....99/100
    sqare(1/2.3/4.5/6.....99/100) < 1/101<1/100

    ReplyDelete

Post a Comment

Popular posts from this blog

Asking a girl out

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this article though :P

Most of my friends already read an article that I wrote more than an year back - "Speak Up"


Here, inspired by the movie, The Beautiful Mind, I give a mathematical analysis of asking a girl out. Nice time it is. Feb 10. No plans for Feb 14 and I am sure this article makes me look even more geekier and all the more reason for me to believe that I will be alone, yet again. But what the hell, lets do it!

Note: This is not an independent analysis. There are many "mathematics sites" which does "similar" analysis.

@Consultants, correct me if I am wrong in my estimates. :P

Why is there a need to be selective?

From the age of 15, I guess there are approximately 3,600 girls I have liked (On average days, I don't see new girls. But going outside, I like about 30 girls. Saying that I go out once every week right …

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process.

1) Calculate P{X>Y}
2) What's the expected value of X
3) What's the expected value of Y

This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment.

Solution:
1)
(Solved by me finally after 13 months :))

Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently.

So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0"

f(State) = P(X>Y | "State" configuration initially)

f(0,0) = 1/4[f(…

Fraction Brainteaser

Source:
Sent to me by Gaurav Sinha

Problem:
Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20 out of 28 Geometry questions. In total, Siddhant scores 25 out of 34. 

Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34.

Note that
a) Vaibhav scores more than Siddhant
b) Siddhant score better than Vaibhav in both individual topics - 5/6 > 20/25 and 20/28 > 6/9

How is it possible?