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

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- shubhamsrg

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

- katieb

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- 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

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

- shubhamsrg

done now..

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- anonymous

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

- anonymous

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

- 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

I hope I made it clear

- shubhamsrg

sorry didnt get it :|

- anonymous

- anonymous

where didnt u understand

- anonymous

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

- anonymous

right? @shubhamsrg

- shubhamsrg

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

- shubhamsrg

yes..following..

- anonymous

Now, take 1 and n and average them

- shubhamsrg

okay,,1+n /2..

- anonymous

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

- shubhamsrg

n/2 ..

- anonymous

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

- shubhamsrg

maybe you meant 2 and n-1 ?

- shubhamsrg

yes,then avg n+1 /2

- anonymous

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

- anonymous

got it?

- shubhamsrg

okay,,i get that part..

- shubhamsrg

buzz ?

- anonymous

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

- anonymous

guys I think I was totally wrong

- anonymous

But got the method to solve this

- shubhamsrg

am all years..

- shubhamsrg

ears **

- shubhamsrg

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

- anonymous

go on saurav

- 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

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

- shubhamsrg

well thats ofcorse not a general soln..

- anonymous

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

- shubhamsrg

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

- anonymous

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

- shubhamsrg

what were you saying about induction?

- 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

hmm

- 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

didnt get the second last step..?

- anonymous

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

- shubhamsrg

i see..

- 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

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

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

- shubhamsrg

yep..

- 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

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

- shubhamsrg

well you can use any legitimate method sir..

- anonymous

no i messed that up, have to try again

- shubhamsrg

people i got the solution!!

- shubhamsrg

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

- 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

Neat solution

- shubhamsrg

thanks! :)

- anonymous

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

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