## shubhamsrg 3 years ago how to prove n! ≤ ( (n+1)/2 )^n for n≥ 1

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

ohh lol..sorry..it should be >= sign,,wait i'll edit the ques..

3. shubhamsrg

done now..

Lemme dust off my real analysis textbook... ahaha...

5. sauravshakya

n! = 1*2*3*...*(n-2)*(n-1)*n Now, (1+n/2) = (2+n-1)/2 = (3+n-2)/2

6. sauravshakya

(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

I hope I made it clear

8. shubhamsrg

sorry didnt get it :|

9. sauravshakya

@shubhamsrg

10. sauravshakya

where didnt u understand

11. sauravshakya

n! = 1*2*3*...*(n-2)*(n-1)*n

12. sauravshakya

right? @shubhamsrg

13. shubhamsrg

so there will be around half number of (1+n)/2 terms..... + conclusion part..

14. shubhamsrg

yes..following..

15. sauravshakya

Now, take 1 and n and average them

16. shubhamsrg

okay,,1+n /2..

17. sauravshakya

So, u will get (n+1)/2 right?

18. shubhamsrg

n/2 ..

19. sauravshakya

u will again get (n+1)/2 right?

20. shubhamsrg

maybe you meant 2 and n-1 ?

21. shubhamsrg

yes,then avg n+1 /2

22. sauravshakya

oh...... 2 and (n-1)

23. sauravshakya

got it?

24. shubhamsrg

okay,,i get that part..

25. shubhamsrg

buzz ?

26. mukushla

check saura's method and also this go by mathematical induction... Use this :$2\le (\frac{n+2}{n+1})^{n+1}$

27. sauravshakya

guys I think I was totally wrong

28. sauravshakya

But got the method to solve this

29. shubhamsrg

am all years..

30. shubhamsrg

ears **

31. shubhamsrg

@mukushla didnt get you.. @sauravshakya your method ?

32. mukushla

go on saurav

33. shubhamsrg

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

take n=2 so n!=2 and (3/2)^2=2.25 so 2<2.25 hence proved

35. shubhamsrg

well thats ofcorse not a general soln..

36. mukushla

u got n! / (n/2) ! > ((n+1)/2 )^(n/2) how can we match this with original inequality?

37. shubhamsrg

you're the expert,,you telll!! :P

38. mukushla

lol...idk if it will help or not.

39. shubhamsrg

what were you saying about induction?

40. mukushla

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

hmm

42. mukushla

$(n+1)!=(n+1).n! \le (n+1).(\frac{n+1}{2})^{n}\le \frac{n+1}{2}(\frac{n+2}{n+1})^{n+1}.(\frac{n+1}{2})^{n}=(\frac{n+2}{2})^{n+1}$

43. shubhamsrg

didnt get the second last step..?

44. mukushla

well $1 \le \frac{1}{2} (\frac{n+2}{n+1})^{n+1}$ but u need to prove this inequality also

45. shubhamsrg

i see..

46. shubhamsrg

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

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

u most prove $$\text{something} \le 0$$ right?

49. shubhamsrg

yep..

50. shubhamsrg

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. satellite73

i did not read all of the above, but is this supposed to be a proof by induction?

52. shubhamsrg

well you can use any legitimate method sir..

53. satellite73

no i messed that up, have to try again

54. shubhamsrg

people i got the solution!!

55. shubhamsrg

it was accidently,,while solving another ques,,and i recalled AM >= GM,,here's my solution

56. shubhamsrg

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

Neat solution

58. shubhamsrg

thanks! :)

59. mukushla

very nice man...$\frac{1+2+...+n}{n}\ge\sqrt[n]{n!}$