Expert: Scott A Wilson - 9/11/2011

Question
QUESTION: the question is attached as gif file

ANSWER: ln(n^2)(Ln(n)-1) < n
That would be the same as 2*ln(n)[ln(n)-1].
If n = e, we have 1(1-0) < e, and that is true.

The function we are dealing with is 2*ln^2(n) - ln(n).
The derivative of that function is 4*ln(n)/n - 1/n = (4*ln(n)-1)/n,
and that is clearly less than 1 when n gets large.

The derivative of n is 1, and that is greater than the one on the left for x large enough.
From, that, we can see that n increases at a faster rate.;


---------- FOLLOW-UP ----------

QUESTION: any idea how to prove this inequality without using calculus?

Answer
It is known that ln(n^2)(ln(n)-1) = 2ln(n)(ln(n)-1).
If n=e, we have 2(1-1) = 0, and that is less than e.
If n=e^2, we have 2*2(2-1) = 4, and that is less than e^2.
If we have n=e^3, we have 2*3(3-1) = 12, and that is less than e^3.
If we have n=e^4, we have 2*4(4-1) = 24, and that is less than e^4.
If we have n=e^5, we have 2*5(5-1) = 40, and that is less than e^5.

See, on the left side of the equation, the amount added can be defined 2x^2 - x.
On the right side, it is increasing powers on the exponential.
If is known that exponentials always incrase faster than polynomials, so this is always true.