The given problem asks to determine whether the series:

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

is convergent or divergent using the integral test.

Steps for applying the Integral Test:

The Integral Test states that for a series $\sum_{n=1}^{\infty} a_n$, if $f(x)$ is a continuous, positive, and decreasing function on $[1, \infty)$ and $a_n = f(n)$, then the behavior of the series (converging or diverging) is the same as the behavior of the integral:

$\int_{1}^{\infty} f(x)\,dx$

For this series, let: $f(x) = \frac{1}{x^3 + x}$ We need to evaluate the improper integral:

$I = \int_{1}^{\infty} \frac{1}{x^3 + x}\, dx$

Simplifying the Integral:

We can factor the denominator: $\frac{1}{x^3 + x} = \frac{1}{x(x^2 + 1)}$

Now, let's evaluate the integral:

$I = \int_{1}^{\infty} \frac{1}{x(x^2 + 1)}\, dx$

This is a standard integral, which can be solved using partial fractions.

We can decompose the fraction as follows: $\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$

Solving for $A$, $B$, and $C$ and then integrating will show whether the integral converges or diverges. If the integral converges, the series converges, and if it diverges, the series diverges.

Let me compute the integral and determine the result.The result of the integral is:

$\int_{1}^{\infty} \frac{1}{x(x^2 + 1)}\, dx = \frac{\log(2)}{2}$

Since $\frac{\log(2)}{2}$ is a finite value, the integral converges. Therefore, by the Integral Test, the given series:

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

converges as well.

Would you like further clarification on the steps? Here are some related questions to deepen your understanding:

1. How does the integral test apply to other types of series?
2. Can the convergence of this series be determined using other methods?
3. What are the conditions for using the integral test?
4. What is the difference between absolute and conditional convergence?
5. How does partial fraction decomposition help in solving integrals?

Tip: Always check that the function used in the integral test is positive, continuous, and decreasing before applying the test.