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shubhamsrg

  • 3 years ago

how to prove n! ≤ ( (n+1)/2 )^n for n≥ 1

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  1. badreferences
    • 3 years ago
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    Maybe I'm wrong, but I was under the impression that factorials are defined only for \([0,\infty)\). So isn't it just a case of demonstrating the conjecture for \(0,1\)? I barely recall reading a paper about evaluating negative factorials with the gamma function, but that was entirely on the complex plane (IIRC).

  2. shubhamsrg
    • 3 years ago
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    ohh lol..sorry..it should be >= sign,,wait i'll edit the ques..

  3. shubhamsrg
    • 3 years ago
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    done now..

  4. badreferences
    • 3 years ago
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    Lemme dust off my real analysis textbook... ahaha...

  5. sauravshakya
    • 3 years ago
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    n! = 1*2*3*...*(n-2)*(n-1)*n Now, (1+n/2) = (2+n-1)/2 = (3+n-2)/2

  6. sauravshakya
    • 3 years ago
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    (1+n/2) = (2+n-1)/2 = (3+n-2)/2 ==== (1+n)/2 Also, n! has n number of terms..... so there will be around half number of (1+n)/2 terms..... and also, ( (n+1)/2 )^n has n number of terms.. Thus, n! ≤ ( (n+1)/2 )^n

  7. sauravshakya
    • 3 years ago
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    I hope I made it clear

  8. shubhamsrg
    • 3 years ago
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    sorry didnt get it :|

  9. sauravshakya
    • 3 years ago
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    @shubhamsrg

  10. sauravshakya
    • 3 years ago
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    where didnt u understand

  11. sauravshakya
    • 3 years ago
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    n! = 1*2*3*...*(n-2)*(n-1)*n

  12. sauravshakya
    • 3 years ago
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    right? @shubhamsrg

  13. shubhamsrg
    • 3 years ago
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    so there will be around half number of (1+n)/2 terms..... + conclusion part..

  14. shubhamsrg
    • 3 years ago
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    yes..following..

  15. sauravshakya
    • 3 years ago
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    Now, take 1 and n and average them

  16. shubhamsrg
    • 3 years ago
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    okay,,1+n /2..

  17. sauravshakya
    • 3 years ago
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    So, u will get (n+1)/2 right?

  18. shubhamsrg
    • 3 years ago
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    n/2 ..

  19. sauravshakya
    • 3 years ago
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    u will again get (n+1)/2 right?

  20. shubhamsrg
    • 3 years ago
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    maybe you meant 2 and n-1 ?

  21. shubhamsrg
    • 3 years ago
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    yes,then avg n+1 /2

  22. sauravshakya
    • 3 years ago
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    oh...... 2 and (n-1)

  23. sauravshakya
    • 3 years ago
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    got it?

  24. shubhamsrg
    • 3 years ago
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    okay,,i get that part..

  25. shubhamsrg
    • 3 years ago
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    buzz ?

  26. mukushla
    • 3 years ago
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    check saura's method and also this go by mathematical induction... Use this :\[2\le (\frac{n+2}{n+1})^{n+1}\]

  27. sauravshakya
    • 3 years ago
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    guys I think I was totally wrong

  28. sauravshakya
    • 3 years ago
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    But got the method to solve this

  29. shubhamsrg
    • 3 years ago
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    am all years..

  30. shubhamsrg
    • 3 years ago
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    ears **

  31. shubhamsrg
    • 3 years ago
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    @mukushla didnt get you.. @sauravshakya your method ?

  32. mukushla
    • 3 years ago
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    go on saurav

  33. shubhamsrg
    • 3 years ago
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    here's what i tried to do.. (n+1)/2 = n/2 + 1/2 now for even n, n/2 + 1 > (n+1)/2 > n/2 (n/2 + 1) (n/2 + 2) > ((n+1)/2)((n+1)/2) > (n/2) (n/2 -1) . . . . (n/2 +1)(n/2 +2)......(n/2 + n/2) > ((n+1)/2 )^(n/2) > (n/2) (n/2 -1).....(n/2 -n/2) n! / (n/2) ! > ((n+1)/2 )^(n/2) > (n/2) ! is this any help? how can we take it further ?

  34. ravimeena
    • 3 years ago
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    take n=2 so n!=2 and (3/2)^2=2.25 so 2<2.25 hence proved

  35. shubhamsrg
    • 3 years ago
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    well thats ofcorse not a general soln..

  36. mukushla
    • 3 years ago
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    u got n! / (n/2) ! > ((n+1)/2 )^(n/2) how can we match this with original inequality?

  37. shubhamsrg
    • 3 years ago
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    you're the expert,,you telll!! :P

  38. mukushla
    • 3 years ago
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    lol...idk if it will help or not.

  39. shubhamsrg
    • 3 years ago
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    what were you saying about induction?

  40. mukushla
    • 3 years ago
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    well for \(n=1\) its true... let suppose\[n! \le (\frac{n+1}{2})^n \]be a true statement...we just need to prove\[(n+1)! \le (\frac{n+2}{2})^{n+1} \]

  41. shubhamsrg
    • 3 years ago
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    hmm

  42. mukushla
    • 3 years ago
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    \[(n+1)!=(n+1).n! \le (n+1).(\frac{n+1}{2})^{n}\le \frac{n+1}{2}(\frac{n+2}{n+1})^{n+1}.(\frac{n+1}{2})^{n}=(\frac{n+2}{2})^{n+1} \]

  43. shubhamsrg
    • 3 years ago
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    didnt get the second last step..?

  44. mukushla
    • 3 years ago
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    well \[1 \le \frac{1}{2} (\frac{n+2}{n+1})^{n+1}\] but u need to prove this inequality also

  45. shubhamsrg
    • 3 years ago
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    i see..

  46. shubhamsrg
    • 3 years ago
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    if we try like this : n = (n+1 /2) + (n-1 /2) n-1 = (n+1 /2) + (n-3 /2) . . . . and multiplying all, n! = (n+1 /2)^n + something.. does this help ??

  47. shubhamsrg
    • 3 years ago
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    RHS will be some polynomial or f(n+1 /2) with roots -(n-1)/2 ,-(n-3)/2 etc.. so there isnt much problem multiplying all..main problem is how to prove!! ?? !!

  48. mukushla
    • 3 years ago
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    u most prove \(\text{something} \le 0\) right?

  49. shubhamsrg
    • 3 years ago
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    yep..

  50. shubhamsrg
    • 3 years ago
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    if n+1 /2 = x, then it'll be of the form x^n + x^(n-1) (0) - x^(n-2)(something +ve) + ............... what will come in ............ ?? am quite sure about what i wrote before ...............

  51. anonymous
    • 3 years ago
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    i did not read all of the above, but is this supposed to be a proof by induction?

  52. shubhamsrg
    • 3 years ago
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    well you can use any legitimate method sir..

  53. anonymous
    • 3 years ago
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    no i messed that up, have to try again

  54. shubhamsrg
    • 3 years ago
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    people i got the solution!!

  55. shubhamsrg
    • 3 years ago
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    it was accidently,,while solving another ques,,and i recalled AM >= GM,,here's my solution

  56. shubhamsrg
    • 3 years ago
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    we know that : [ (1 + 2 +3 ..... n)/n ] ^n >= 1.2.3......n or [ n(n+1)/2n ]^n >= n! => (n+1 /2)^n >=n! lol...

  57. mukushla
    • 3 years ago
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    Neat solution

  58. shubhamsrg
    • 3 years ago
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    thanks! :)

  59. mukushla
    • 3 years ago
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    very nice man...\[\frac{1+2+...+n}{n}\ge\sqrt[n]{n!}\]

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