The series provided is:

$\sum_{n=1}^{\infty} \frac{\ln n}{n^3}$

To apply the integral test, we first convert the sum into an integral. The integral test is applicable if the corresponding function $f(x) = \frac{\ln x}{x^3}$ is positive, continuous, and decreasing for $x \geq 1$.

Now, we analyze the integral:

$I = \int_1^{\infty} \frac{\ln x}{x^3} \, dx$

Step 1: Integration by parts

We can solve this using integration by parts. Let:

• $u = \ln x$ (which implies $du = \frac{1}{x} dx$),
• $dv = \frac{1}{x^3} dx$ (which gives $v = -\frac{1}{2x^2}$).

Now, applying the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

This becomes:

$I = \left[ \ln x \cdot \left( -\frac{1}{2x^2} \right) \right]_1^{\infty} - \int_1^{\infty} \left( -\frac{1}{2x^2} \right) \cdot \frac{1}{x} \, dx$

Step 2: Evaluate the boundary terms

For the first term:

$\lim_{x \to \infty} \left( \ln x \cdot \frac{1}{x^2} \right) = 0 \quad \text{and at} \quad x = 1, \quad \ln 1 = 0$

Thus, the boundary terms vanish, and we are left with the integral:

$I = \frac{1}{2} \int_1^{\infty} \frac{1}{x^3} \, dx$

Step 3: Solve the remaining integral

This is a straightforward integral:

$\int_1^{\infty} \frac{1}{x^3} \, dx = \left[ -\frac{1}{2x^2} \right]_1^{\infty} = 0 - \left( -\frac{1}{2} \right) = \frac{1}{2}$

Thus, the value of the original integral is:

$I = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

Conclusion:

Since the integral converges to a finite value, by the integral test, the series $\sum_{n=1}^{\infty} \frac{\ln n}{n^3}$ converges.

Would you like further details on the integral test or the method used?

Here are five related questions for further exploration:

1. Can the ratio test be applied to this series for comparison?
2. How does the behavior of logarithmic terms influence the convergence of series?
3. What is the difference between the integral test and the comparison test?
4. Can the limit comparison test confirm the result of the integral test in this case?
5. How do we determine the rate of convergence for such series?

Tip: Always check if the function $f(x)$ is decreasing before applying the integral test to ensure the method is valid.