Show this inequality for integer numbers (difficult problem!)?

Question

http://m1.freeshare.us/155fs152333.gif

Answer 1

I’m going to reduce the bulkiness of this problem by substituting m = 2n – 1, where it is understood that m is odd.
We are required to prove that
(m/e)^(m/2) < 1*3*5*…*m < [(m+2)/e]^[(m+2)/2]

Taking logs of all three quantities:
(m/2)*[ln(m)–1] < ln(1)+ln(3) +…+ ln(m) < (m+2)/2 *[ ln(m+2)–1]

Let f(m) = (m/2)*[ln(m) – 1]
Let g(m) = (m+2)/2 * [ ln(m+2) – 1]

We need to prove, for all odd values of m, that
A) f(m) < ln(1) + ln(3) + … + ln(m)
B) g(m) > ln(1) + ln(3) + … + ln(m)

PROOF OF STATEMENT (A) BY INDUCTION:
We can verify that both hold true for m = 1
Then we assume that both hold true for m = k
Now we are required to prove they hold true for m = k+2
ie we must prove that
f(k+2) < ln(1) + ln(3) + … + ln(k) + ln(k+2)
given that f(k) < ln(1) + ln(3) + … + ln(k)
thus we must prove f(k+2) – f(k) < ln(k+2)

After simplifying, this reduces to proving that
k/2 * ln(1 + 2/k) < 1
or 1 + 2/k < e^(2/k)

This can be accomplished by comparing the LHS with the Maclaurin series of the RHS
1 + 2/k < 1 + 2/k + (2/k)^2 / 2 + …
Thus statement A holds true for all m.

PROOF OF STATEMENT (B) BY INDUCTION:
Again statement B holds true for m = 1, and using similar reasoning to above, we end up having to prove that
g(k+2) – g(k) > ln(k+2)

After simplifying, this reduces to proving that
ln(1 + x) > x / (1 + x)
or (1+x)*ln(1+x) > x
where x = 2/(k + 2), thus 0 < x < 2/3 for k >= 1

Using Maclaurin series, we can write the LHS as
x + ∑ {n = 2 to inf} of (-1)^n / (n^2 -n) * x^n

All we have to do at this point is show that
{ ∑ {n = 2 to inf} of (-1)^n / (n^2 -n) * x^n } > 0

This series consist of alternating positive and negative terms, so it is sufficient to show that each successive pair of terms sums to a positive value. Consider terms 2n and 2n+1.

x^2n / [2n(2n-1)] – x^(2n+1) / [2n(2n+1)]
= x^2n / [2n(2n-1)(2n+1)] * [(2n+1) – x*(2n-1)]
This is only negative if
[(2n+1) – x*(2n-1)] < 0
or if x > (2n+1) / (2n-1) > 1
But since x < 2/3, that summation must be positive, hence we can conclude that statement B is true for all m.

THIS COMPLETES THE PROOF BY INDUCTION.

Answer 2

Just a suggestion, I will work on it later (sorry, I have a headache now) , but have you tried mathematical induction?

EDIT: okay, unless I made a mistake I have showed the inequality using mathematical induction but the solution is quite long to write here. Maybe i did not see an easier way, but within my inequality, I found I came across another inequality which I used another application of mathematical induction in.
Try mathematical induction if you like.
I will see if I can use cheeser's suggestion.
I am sure that some smart people out there will have a more
elegant solution than the one I did.

Answer 3

Adopting Dr. D’s notation m = 2n-1, here is another way to prove the inequality
(m/2) ln (m) – n + ½ < ln1 + ln 3 + … + ln m.
Since ln(2x-1) is an increasing function,
Integral{ln(2x-1); x = 1 to n} < ln 1 + ln 3 + … + ln m
(It may help to sketch the function ln(2x-1) and draw the over-estimating rectangles for the integral in order to see this.)
The LHS integral evaluates to (m/2) ln (m) – n + 1. Since
(m/2) ln (m) – n + ½ < (m/2) ln (m) – n + 1, we actually have proved a tighter inequality than the problem requires, i.e. we have shown
(m/2) ln (m) – n + 1 < ln 1 + ln 3 + … + ln m.
The remaining inequality of the problem,
ln 1 + ln 3 + … + ln m < (2n+1)/2 * ln ((2n+1)/2) – n -1/2,
can be proved similarly through consideration of the integral from 1 to n+1 of ln(2x-1) and comparing it with the under-estimating rectangles.

Answer 4

This is a bit tricky. It follows from the following two facts:

http://planetmath.org/encyclopedia/AsymptoticBoundsForFactorial.html

Those are general asymptotic bounds for the factorial function. However, what you have IS a factorial function, since the thing in the middle is the same as:


(2n)! / ( n! 2^n)

Using the bounds in the first with the expression above, it's a matter of plugging it in and manipulating a few powers. That's all it takes. I hope that's sufficient, I don't have time this weekend to get into explaining things in any more depth than that.