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shubhamsrg Group Title

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

  • 2 years ago
  • 2 years ago

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  1. badreferences Group Title
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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 years ago
  2. shubhamsrg Group Title
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    ohh lol..sorry..it should be >= sign,,wait i'll edit the ques..

    • 2 years ago
  3. shubhamsrg Group Title
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    done now..

    • 2 years ago
  4. badreferences Group Title
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    Lemme dust off my real analysis textbook... ahaha...

    • 2 years ago
  5. sauravshakya Group Title
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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

    • 2 years ago
  6. sauravshakya Group Title
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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

    • 2 years ago
  7. sauravshakya Group Title
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    I hope I made it clear

    • 2 years ago
  8. shubhamsrg Group Title
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    sorry didnt get it :|

    • 2 years ago
  9. sauravshakya Group Title
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    @shubhamsrg

    • 2 years ago
  10. sauravshakya Group Title
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    where didnt u understand

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

    • 2 years ago
  12. sauravshakya Group Title
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    right? @shubhamsrg

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

    • 2 years ago
  14. shubhamsrg Group Title
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    yes..following..

    • 2 years ago
  15. sauravshakya Group Title
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    Now, take 1 and n and average them

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

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

    • 2 years ago
  18. shubhamsrg Group Title
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    n/2 ..

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

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

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

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

    • 2 years ago
  23. sauravshakya Group Title
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    got it?

    • 2 years ago
  24. shubhamsrg Group Title
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    okay,,i get that part..

    • 2 years ago
  25. shubhamsrg Group Title
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    buzz ?

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

    • 2 years ago
  27. sauravshakya Group Title
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    guys I think I was totally wrong

    • 2 years ago
  28. sauravshakya Group Title
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    But got the method to solve this

    • 2 years ago
  29. shubhamsrg Group Title
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    am all years..

    • 2 years ago
  30. shubhamsrg Group Title
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    ears **

    • 2 years ago
  31. shubhamsrg Group Title
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    @mukushla didnt get you.. @sauravshakya your method ?

    • 2 years ago
  32. mukushla Group Title
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    go on saurav

    • 2 years ago
  33. shubhamsrg Group Title
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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 ?

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

    • 2 years ago
  35. shubhamsrg Group Title
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    well thats ofcorse not a general soln..

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

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

    • 2 years ago
  38. mukushla Group Title
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    lol...idk if it will help or not.

    • 2 years ago
  39. shubhamsrg Group Title
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    what were you saying about induction?

    • 2 years ago
  40. mukushla Group Title
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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\[(n+1)! \le (\frac{n+2}{2})^{n+1} \]

    • 2 years ago
  41. shubhamsrg Group Title
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    hmm

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

    • 2 years ago
  43. shubhamsrg Group Title
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    didnt get the second last step..?

    • 2 years ago
  44. mukushla Group Title
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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

    • 2 years ago
  45. shubhamsrg Group Title
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    i see..

    • 2 years ago
  46. shubhamsrg Group Title
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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 ??

    • 2 years ago
  47. shubhamsrg Group Title
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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!! ?? !!

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

    • 2 years ago
  49. shubhamsrg Group Title
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    yep..

    • 2 years ago
  50. shubhamsrg Group Title
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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 ...............

    • 2 years ago
  51. satellite73 Group Title
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    i did not read all of the above, but is this supposed to be a proof by induction?

    • 2 years ago
  52. shubhamsrg Group Title
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    well you can use any legitimate method sir..

    • 2 years ago
  53. satellite73 Group Title
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    no i messed that up, have to try again

    • 2 years ago
  54. shubhamsrg Group Title
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    people i got the solution!!

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

    • 2 years ago
  56. shubhamsrg Group Title
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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...

    • 2 years ago
  57. mukushla Group Title
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    Neat solution

    • 2 years ago
  58. shubhamsrg Group Title
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    thanks! :)

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

    • 2 years ago
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