Let us solve the two parts of the problem using the Integral Test to determine whether the series converges or diverges.

Part (a):

The series is:

$\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$

Step 1: Define the corresponding function

Let $f(x) = \frac{x}{x^2 + 1}$. The function $f(x)$ is:

• Continuous for $x > 0$,
• Positive for $x > 0$,
• Decreasing for $x > 0$ (we can verify this by checking the derivative).

Step 2: Compute the improper integral

We evaluate:

$\int_{1}^{\infty} \frac{x}{x^2 + 1} \, dx$

Use substitution: let $u = x^2 + 1$, so $du = 2x \, dx$.

$\int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(x^2 + 1) + C$

Now, compute the definite integral:

$\int_{1}^{\infty} \frac{x}{x^2 + 1} \, dx = \lim_{b \to \infty} \left[ \frac{1}{2} \ln(x^2 + 1) \right]_{1}^{b}$

$= \lim_{b \to \infty} \left( \frac{1}{2} \ln(b^2 + 1) - \frac{1}{2} \ln(1^2 + 1) \right)$

As $b \to \infty$, $\ln(b^2 + 1) \to \infty$. Therefore, the integral diverges.

Conclusion:

The integral diverges, so by the Integral Test, the series $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ diverges.

Part (b):

The series is:

$\sum_{n=1}^{\infty} n^2 e^{-n^3}$

Step 1: Define the corresponding function

Let $f(x) = x^2 e^{-x^3}$. The function $f(x)$ is:

• Continuous for $x > 0$,
• Positive for $x > 0$,
• Decreasing for $x > 0$ (we can verify this by checking the derivative).

Step 2: Compute the improper integral

We evaluate:

$\int_{1}^{\infty} x^2 e^{-x^3} \, dx$

Use substitution: let $u = x^3$, so $du = 3x^2 \, dx$, and $x^2 \, dx = \frac{1}{3} du$.

$\int x^2 e^{-x^3} \, dx = \frac{1}{3} \int e^{-u} \, du = \frac{1}{3} (-e^{-u}) + C = -\frac{1}{3} e^{-x^3} + C$

Now, compute the definite integral:

$\int_{1}^{\infty} x^2 e^{-x^3} \, dx = \lim_{b \to \infty} \left[ -\frac{1}{3} e^{-x^3} \right]_{1}^{b}$

$= \lim_{b \to \infty} \left( -\frac{1}{3} e^{-b^3} + \frac{1}{3} e^{-1} \right)$

As $b \to \infty$, $e^{-b^3} \to 0$. Therefore:

$\int_{1}^{\infty} x^2 e^{-x^3} \, dx = \frac{1}{3} e^{-1}$

This is a finite value.

Conclusion:

The integral converges, so by the Integral Test, the series $\sum_{n=1}^{\infty} n^2 e^{-n^3}$ converges.

Final Answers:

• (a) The series $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$ diverges.
• (b) The series $\sum_{n=1}^{\infty} n^2 e^{-n^3}$ converges.

Would you like further clarifications or step-by-step details?

Related Questions:

1. What are the general conditions for applying the Integral Test?
2. How does the comparison of a series to its integral ensure convergence or divergence?
3. What is the significance of the substitution method in improper integrals?
4. Can the Integral Test fail, and if so, under what circumstances?
5. How does the behavior of $e^{-x}$ contribute to convergence in series?

Tip:

Always verify that the function used in the Integral Test satisfies the conditions: positive, continuous, and decreasing. If any condition fails, the test cannot be applied.