Prove Big-Omega

Question

Question:

(5n^2)(ln(n)) is big-omega of n(ln(n)^2)

What I have tried:

Exist c > 0, n0 > 0

(5n^2)(ln(n)) >= cn(ln(n)^2) for all n >= n0

(5n^2)(ln(n)) >= n(ln(n)) (for n >= 1) >= n(ln(n)^2) (for n <= 1)

so this concludes that when n = 1 = n0, (5n^2)(ln(n)) is big-omega of n(ln(n)^2); but this does not meet the requirement of (for all n >= n0).

I stuck here and can anyone help?

Answer 1

My first thought:

if

 (5n^2)(ln(n)) is big omega of n(ln(n)^2)

then

 (5n) is big omega of ln(n)

which is fundamental. Look;

exists
    c = 1 and n0 = 1,
such that
    5n >= ln(n); for all n >= n0

Expanding the series for first few elements gives:

 -------------------------
|   n   |   5n   | ln(n)  |
|-------|--------|--------|
|    1  |     5  |  0.00  |
|    2  |    10  |  0.69  |
|    3  |    15  |  1.10  |
|    4  |    20  |  1.39  |
|    5  |    25  |  1.61  |
|   10  |    50  |  2.30  |
|  100  |   500  |  4.61  |
| 1000  |  5000  |  6.91  |
 -------------------------