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

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

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Other answers:

Lemme dust off my real analysis textbook... ahaha...
n! = 1*2*3*...*(n-2)*(n-1)*n Now, (1+n/2) = (2+n-1)/2 = (3+n-2)/2
(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
I hope I made it clear
sorry didnt get it :|
where didnt u understand
n! = 1*2*3*...*(n-2)*(n-1)*n
so there will be around half number of (1+n)/2 terms..... + conclusion part..
yes..following..
Now, take 1 and n and average them
okay,,1+n /2..
So, u will get (n+1)/2 right?
n/2 ..
u will again get (n+1)/2 right?
maybe you meant 2 and n-1 ?
yes,then avg n+1 /2
oh...... 2 and (n-1)
got it?
okay,,i get that part..
buzz ?
check saura's method and also this go by mathematical induction... Use this :\[2\le (\frac{n+2}{n+1})^{n+1}\]
guys I think I was totally wrong
But got the method to solve this
am all years..
ears **
@mukushla didnt get you.. @sauravshakya your method ?
go on saurav
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 ?
take n=2 so n!=2 and (3/2)^2=2.25 so 2<2.25 hence proved
well thats ofcorse not a general soln..
u got n! / (n/2) ! > ((n+1)/2 )^(n/2) how can we match this with original inequality?
you're the expert,,you telll!! :P
lol...idk if it will help or not.
what were you saying about induction?
well for \(n=1\) its true... let suppose\[n! \le (\frac{n+1}{2})^n \]be a true statement...we just need to prove\[(n+1)! \le (\frac{n+2}{2})^{n+1} \]
hmm
\[(n+1)!=(n+1).n! \le (n+1).(\frac{n+1}{2})^{n}\le \frac{n+1}{2}(\frac{n+2}{n+1})^{n+1}.(\frac{n+1}{2})^{n}=(\frac{n+2}{2})^{n+1} \]
didnt get the second last step..?
well \[1 \le \frac{1}{2} (\frac{n+2}{n+1})^{n+1}\] but u need to prove this inequality also
i see..
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 ??
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!! ?? !!
u most prove \(\text{something} \le 0\) right?
yep..
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 ...............
i did not read all of the above, but is this supposed to be a proof by induction?
well you can use any legitimate method sir..
no i messed that up, have to try again
people i got the solution!!
it was accidently,,while solving another ques,,and i recalled AM >= GM,,here's my solution
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...
Neat solution
thanks! :)
very nice man...\[\frac{1+2+...+n}{n}\ge\sqrt[n]{n!}\]

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