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badreferencesBest ResponseYou've already chosen the best response.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).
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
ohh lol..sorry..it should be >= sign,,wait i'll edit the ques..
 one year ago

badreferencesBest ResponseYou've already chosen the best response.1
Lemme dust off my real analysis textbook... ahaha...
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
n! = 1*2*3*...*(n2)*(n1)*n Now, (1+n/2) = (2+n1)/2 = (3+n2)/2
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
(1+n/2) = (2+n1)/2 = (3+n2)/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
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
I hope I made it clear
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
sorry didnt get it :
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
where didnt u understand
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
n! = 1*2*3*...*(n2)*(n1)*n
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
right? @shubhamsrg
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
so there will be around half number of (1+n)/2 terms..... + conclusion part..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Now, take 1 and n and average them
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
So, u will get (n+1)/2 right?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
u will again get (n+1)/2 right?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
maybe you meant 2 and n1 ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
yes,then avg n+1 /2
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
oh...... 2 and (n1)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
okay,,i get that part..
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
check saura's method and also this go by mathematical induction... Use this :\[2\le (\frac{n+2}{n+1})^{n+1}\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
guys I think I was totally wrong
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
But got the method to solve this
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
@mukushla didnt get you.. @sauravshakya your method ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
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 ?
 one year ago

ravimeenaBest ResponseYou've already chosen the best response.0
take n=2 so n!=2 and (3/2)^2=2.25 so 2<2.25 hence proved
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
well thats ofcorse not a general soln..
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
u got n! / (n/2) ! > ((n+1)/2 )^(n/2) how can we match this with original inequality?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
you're the expert,,you telll!! :P
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
lol...idk if it will help or not.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
what were you saying about induction?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
well for \(n=1\) its true... let suppose\[n! \le (\frac{n+1}{2})^n \]be a true statement...we just need to prove\[(n+1)! \le (\frac{n+2}{2})^{n+1} \]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
\[(n+1)!=(n+1).n! \le (n+1).(\frac{n+1}{2})^{n}\le \frac{n+1}{2}(\frac{n+2}{n+1})^{n+1}.(\frac{n+1}{2})^{n}=(\frac{n+2}{2})^{n+1} \]
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
didnt get the second last step..?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
well \[1 \le \frac{1}{2} (\frac{n+2}{n+1})^{n+1}\] but u need to prove this inequality also
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
if we try like this : n = (n+1 /2) + (n1 /2) n1 = (n+1 /2) + (n3 /2) . . . . and multiplying all, n! = (n+1 /2)^n + something.. does this help ??
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
RHS will be some polynomial or f(n+1 /2) with roots (n1)/2 ,(n3)/2 etc.. so there isnt much problem multiplying all..main problem is how to prove!! ?? !!
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
u most prove \(\text{something} \le 0\) right?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
if n+1 /2 = x, then it'll be of the form x^n + x^(n1) (0)  x^(n2)(something +ve) + ............... what will come in ............ ?? am quite sure about what i wrote before ...............
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i did not read all of the above, but is this supposed to be a proof by induction?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
well you can use any legitimate method sir..
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
no i messed that up, have to try again
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
people i got the solution!!
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
it was accidently,,while solving another ques,,and i recalled AM >= GM,,here's my solution
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.2
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...
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
very nice man...\[\frac{1+2+...+n}{n}\ge\sqrt[n]{n!}\]
 one year ago
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