shubhamsrg
  • shubhamsrg
how to prove n! ≤ ( (n+1)/2 )^n for n≥ 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
  • shubhamsrg
ohh lol..sorry..it should be >= sign,,wait i'll edit the ques..
shubhamsrg
  • shubhamsrg
done now..

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

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