The given problem asks for the values of $q$ for which the series

$\sum_{n=4}^{\infty} \frac{3}{n^q \ln n}$

converges. To determine the convergence of this series, we need to analyze it using standard convergence tests.

Step 1: Analyze the Series

The general term of the series is:

$a_n = \frac{3}{n^q \ln n}.$

For large $n$, this behaves like:

$a_n \approx \frac{1}{n^q \ln n}.$

The presence of $\ln n$ complicates things slightly, but we can use a comparison test with a simpler series $\sum \frac{1}{n^p}$, which converges when $p > 1$ (this is the p-series).

Step 2: Compare with a Simplified Series

For large $n$, the logarithmic term $\ln n$ grows slowly compared to powers of $n$, so the convergence behavior is largely dictated by the power of $n$. Specifically:

• If $q > 1$, the terms $n^q \ln n$ grow rapidly, making the terms of the series shrink fast enough for convergence.
• If $q \leq 1$, the series does not converge because the terms shrink too slowly (or not at all for small $q$).

Thus, for the series to converge, the exponent $q$ must be greater than 1.

Conclusion:

The series converges when $q > 1$. Therefore, the correct answer is option 5: $q > 1$.

Would you like more details or have any further questions? Here are some related questions for further exploration:

1. How does the logarithmic term $\ln n$ affect convergence in other series?
2. What happens if $q = 1$? Does the series converge or diverge?
3. Can we apply the ratio test or root test to this series for convergence?
4. How would the series change if the lower bound of summation were different, say $n = 1$?
5. What is the general behavior of p-series and when do they converge?

Tip: Always simplify a complex series by comparing it to well-known convergent or divergent series like the p-series to make analysis easier.