Let's start with a simple idea: an ordered list of numbers. This is a sequence. Each number in the list is called a term, and we can often describe the entire sequence with a formula for its $n$-th term, usually written as $a_n$. For example, the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...$ can be described by the formula $a_n = \frac{1}{n}$ for $n \ge 1$.

The key question for a sequence is whether it converges or diverges. A sequence converges if its terms get closer and closer to a single, finite number as $n$ approaches infinity. We call this number the limit. In our example, as $n$ gets huge, $\frac{1}{n}$ gets very close to 0. So, we say the sequence converges to 0.

If a sequence doesn't approach a specific finite number, it diverges. This can happen if the terms grow without bound (like $a_n = n^2$) or oscillate (like $a_n = (-1)^n$).

A series is what you get when you add up all the terms of a sequence. We write this using sigma notation:

The main question for a series is the same: does it converge or diverge? A series converges if the sum of its terms approaches a finite number. Otherwise, it diverges. This is a much trickier question than it is for sequences. Just because the terms of a sequence approach zero doesn't guarantee that their sum will be finite.

To figure out if a series converges, we have a toolkit of tests. It's often a process of elimination, and recognizing the form of the series is key to choosing the right tool.

Let's walk through the most common tests.

This should always be your first step. It's a quick check. If the terms you're adding don't even go to zero, there's no way their sum can be finite.

Rule: If $\lim_{n \to \infty} a_n \neq 0$ or the limit doesn't exist, then $\sum a_n$ diverges.

Crucially, if the limit is zero, this test tells you nothing. The series might converge or it might diverge (like the harmonic series). You must use another test.

This test connects series to the integrals we've already studied. Suppose there's a function $f(x)$ that corresponds to the terms of your series, so $f(n) = a_n$. If this function is continuous, positive, and decreasing for $x \ge 1$, then the series $\sum a_n$ and the improper integral $\int_1^\infty f(x) dx$ either both converge or both diverge.

This is the formal way to prove that the harmonic series diverges, by comparing it to $\int_1^\infty \frac{1}{x} dx$, which we know diverges.

The idea here is simple. If you have a series with positive terms, you can compare it to a known series.

1. If your series is term-by-term smaller than a known convergent series, your series must also converge.
2. If your series is term-by-term larger than a known divergent series, your series must also diverge.

For example, to test $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$, we can see that $\frac{1}{n^2 + 1} < \frac{1}{n^2}$. We know that $\sum \frac{1}{n^2}$ is a convergent p-series. Since our series is smaller, it must also converge.

What about series with negative terms, like $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \dots$? This is the alternating harmonic series. For these, we have a special test. An alternating series converges if two conditions are met:

1. The terms (ignoring the sign) are decreasing. So, $b_{n+1} \le b_n$.
2. The limit of the terms (ignoring the sign) is zero. So, $\lim_{n \to \infty} b_n = 0$.

The alternating harmonic series meets both conditions, so it converges.

The final two tests are powerful, especially for series involving factorials or $n$-th powers.

If $L < 1$, the series converges absolutely. If $L > 1$, the series diverges. If $L = 1$, the test is inconclusive.

The conclusion is the same as the ratio test. If $L < 1$, it converges. If $L > 1$, it diverges. If $L = 1$, the test fails and you need another method.

Choosing the right test takes practice. Start by looking at the form of $a_n$. Does it alternate? Is it a rational function? Does it have factorials or powers of $n$? Answering these questions will point you toward the most effective test.