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