Solve mathematically AND and try to explain in English words what is conceptuall

Question

Solve mathematically AND and try to explain in English words what is conceptually going on throughout at least a couple of the steps. So, I need the mathematical steps and some English words as well. Be sure to state what tests you use etc.

Problem 3 If n is sufficiently large, the following functions of n can be arranged in an increasing order, so that each function is very much larger than the one preceding it. List the functions below in order of size from smallest to largest. n; nn ; lnn ; 4n ; 2n ; n ln n ; 2(n2) ; ; (n3 + 1)2/3. Where would n! fit in this list?

Explanation / Answer

ln n; n; n ln n; (n^3+1)^(2/3); sqrt(n^6+1); 2^n; 4^n; n^n; 2^(n^2)

To show that this is correct, we will make pairwise comparisons as we go up the ladder.

The derivative of ln n is 1/n and the derivative of n is 1, so clearly n increases faster, and at 1, n is 1 and ln n is 0, so n > ln n for all n

for all n >= 3, ln n > 1, so n ln n > n

(n^3+1)^2/3 goes to n^(3*2/3) = n^2, so this is greater than n ln n (the comparison is equivalent to n versus ln n; just multiply each side by n)

To show it goes to n^2, clearly, it is greater than (n^3)^2/3 = n^2 for all n, and

(n+1/(3n^2))^3 = n^3 + 1 + 1/3n^3 + 1/27n^6 > n^3 + 1, and ((n+1/(3n^2))^3)^(2/3) = (n+1/(3n^2))^2 = n^2 + 2/3n + 1/9n^4 has limit n^2, so n^2 <= lim n-> inf. (n^3+1)^2/3 <= n^2, so lim n-> inf. (n^3+1)^2/3 = n^2

sqrt(n^6+1) goes to n^(6*1/2) = n^3, so this is greater than n^2 or the previous term

(To easily show it goes to n^3, it is clearly greater than n^3 for all n, and (n^3 + 1/2n^3)^2 = (n^6 + 1 + 1/4n^6) > n^6 + 1, and sqrt((n^3 + 1/2n^3)^2) = n^3 + 1/2n^3, which has limit n^3.

We may show that 2^n > n^3 for all n >= 10

At 10, 2^10 = 1024

At 10, 10^3 = 1000

Then, the ratio is 2 between successive terms for 2^n. For (n+1)^3/n^3 = (1+1/n)^3, the ratio is constantly decreasing and goes to 1. At 10, it equals 1.1^3 = 1.331 < 2. As the ratio is always < 2 for n >= 10 and 2^10 > 10^3, 2^n > n^3 for all n >= 10

Then, 4^n = (2^2)^n = 2^2n > 2^n

As n > 2, n^n > 2^n (The ratio between terms of the two series is (n/2)^n is > 1)

2^(n^2) = (2^n)^n and 2^n > n, so (2^n)^n > n^n

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