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.

An improper integral is convergent if it evaluates to a finite number.

An improper integral is divergent if it evaluates to $\pm\infty$ or does not exist.

A function $f(x)$ is positive and decreasing over an interval if $f(x) > 0$ and $f'(x) < 0$ for all $x$ in that interval.

$a_n = f(n)$ means the terms of the series are obtained by evaluating the function $f(x)$ at integer values $n$.

A method to determine the convergence or divergence of an infinite series by comparing it to an improper integral.

The sum of an infinite number of terms.

An integral with infinite limits of integration or a discontinuous integrand.

The sum of the infinite series approaches a finite value.

The sum of the infinite series does not approach a finite value.

A technique for evaluating integrals by substituting a function $u$ for part of the integrand.

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 $du = g'(x) , dx$

$\int c , dx = cx + C$

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

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

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