The function $f(x)$ must be continuous, positive, and decreasing on the interval $[k, \infty)$, and $a_n = f(n)$.

If the function is not positive, the comparison between the integral and the series is not valid, as the areas and sums could have different signs.

If the function is not decreasing, the integral may not accurately represent the sum of the series, as the terms may not consistently get smaller.

If the improper integral converges, the corresponding infinite series also converges. If the improper integral diverges, the corresponding infinite series also diverges.

The limit of the integral as the upper bound approaches infinity exists and is a finite number.

The limit of the integral as the upper bound approaches infinity does not exist or is infinite.

U-substitution is a technique used to simplify the integral, making it easier to evaluate and determine its convergence or divergence.

The starting index 'k' determines the lower limit of integration and the first term of the series. The conditions of the Integral Test must hold for $x \geq k$.

The integral test cannot be applied, and another test for convergence or divergence must be used.

To determine whether an infinite series converges or diverges by comparing it to an improper integral.

1. Verify $f(x) = \frac{1}{x^2+1}$ is positive, continuous, and decreasing for $x \geq 1$. 2. Evaluate $\int_1^{\infty} \frac{1}{x^2+1} , dx$. 3. If the integral converges, the series converges; if it diverges, the series diverges.

1. Use u-substitution: $u = \ln(x)$, $du = \frac{1}{x} , dx$. 2. Rewrite the integral as $\int \frac{1}{u} , du$. 3. Evaluate to get $\ln|u| + C = \ln|\ln(x)| + C$. 4. Evaluate the improper integral.

Show that the derivative $f'(x) = -\frac{1}{x^2}$ is negative for $x > 0$.

1. Verify $f(x)$ is positive, continuous, and decreasing for $x \geq k$. 2. Evaluate the improper integral $\int_k^{\infty} f(x) , dx$. 3. Determine if the integral converges or diverges. 4. Conclude the series converges/diverges accordingly.

Choose $u = \ln(n)$ because its derivative $\frac{1}{n}$ is also present in the series/integral.

Compute the derivative of the function. If the derivative is negative over the interval of interest, the function is decreasing.

Replace the infinite limit with a variable, say $b$, and take the limit as $b$ approaches infinity.

Apply the integral test with $f(x) = \frac{1}{x^p}$. The series converges if $p > 1$ and diverges if $p \leq 1$.

Look for a function $f(x)$ such that $f(n)$ matches the terms of the series and $f(x)$ is positive, continuous, and decreasing.

Ensure that the function $f(x)$ satisfies the conditions of the Integral Test for all $x$ greater than or equal to the lower limit.

If $f(x)$ is positive, continuous, and decreasing on $[k, \infty)$ and $a_n = f(n)$, then $\sum_{n=k}^{\infty} a_n$ converges if and only if $\int_k^{\infty} f(x) , dx$ converges.

If $\int_k^{\infty} f(x) , dx$ converges, then $\sum_{n=k}^{\infty} a_n$ converges.

If $\int_k^{\infty} f(x) , dx$ diverges, then $\sum_{n=k}^{\infty} a_n$ diverges.

$\int \frac{1}{x} , dx = \ln|x| + C$

$\int \frac{1}{1+x^2} , dx = \arctan(x) + C$

$\int f(g(x))g'(x) , dx = \int f(u) , du$, where $u = g(x)$ and $du = g'(x) , dx$

$\int c , dx = cx + C$

$\int f'(g(x))g'(x)dx = f(g(x)) + C$

$\int \frac{1}{a^2 + x^2} dx = \frac{1}{a}arctan(\frac{x}{a}) + C$

$\int \frac{1}{x} dx = ln|x| + C$