Final answer:
The convergence of the series Σ (log_n(n!))/(n³) can be assessed with a comparison to a convergent p-series. Using Stirling's approximation suggests this series is similar to the convergent series Σ 1/n², pointing towards convergence, though a rigorous comparison test is needed for a definitive conclusion.
Step-by-step explanation:
The question asks to determine whether the series Σ from n=2 to infinity of (log_n(n!))/(n³) is convergent or divergent. To analyze the convergence of the series, one can apply a comparison test, which involves comparing the terms of the given series with a series that is known to converge or diverge. A helpful comparison might be to the series Σ 1/n², which is a p-series with p = 2 and is therefore known to be convergent.
Using Stirling's approximation for n!, n! ≈ √(2πn)(n/e)^n, we can say that for large n, log_n(n!) ≈ log_n(n^n) = n. Hence, the term (log_n(n!))/(n³) will approximate to 1/n², suggesting that the given series is similar to a convergent p-series. Keeping in mind that this is an informal argument and would require a more rigorous proof for full validation, one might conclude that the series appears to be convergent.