Helpful context:

Achilles is racing a tortoise. He is ten times faster. The tortoise gets a 100-meter head start.

Achilles runs 100 meters. The tortoise has moved 10 meters. Achilles runs those 10 meters. The tortoise has moved 1 more meter. He runs that 1 meter. The tortoise has moved 10 centimeters. And so on, forever.

Zeno’s argument: before Achilles can pass the tortoise, he must first reach wherever the tortoise currently is. But by the time he gets there, the tortoise has moved again. There is always one more step. An infinite number of steps. Therefore, Achilles can never catch the tortoise.

And yet. In real life, Achilles catches the tortoise after about 11.1 seconds. The paradox is not a paradox about physics. It is a question about mathematics: can you add up infinitely many positive numbers and get a finite answer?

The ancient Greeks were disturbed by this question. They had no tools to answer it. We do. The answer is yes - you can. And that yes is the fundamental insight of this entire subject.

Here is the sum Zeno is computing: Achilles runs 100 meters, then 10, then 1, then 0.1, then 0.01, and so on forever. The total distance is:

$$100 + 10 + 1 + 0.1 + 0.01 + \cdots$$

Factor out 100:

$$100 \cdot \left(1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \cdots\right)$$

This inner sum equals $\frac{1}{1 - 1/10} = \frac{10}{9}$. So the total is $100 \cdot \frac{10}{9} = \frac{1000}{9} \approx 111.1$ meters. Finite. Achilles does catch the tortoise. An infinite sum gave a finite answer.

How? Why? When does this happen and when does it not? Those are the questions this post answers. They turn out to be subtle, important, and genuinely hard. Let us start from the beginning.

Section 1: What Is a Sequence?

Before we can add infinitely many numbers, we need to understand an infinite list of numbers.

A sequence is a function from the natural numbers to the real numbers: for each $n = 1, 2, 3, \ldots$, there is a term $a_n$. We write the sequence as $(a_n)$ or $(a_1, a_2, a_3, \ldots)$.

Examples:

$$a_n = \frac{1}{n}: \quad 1,\ \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4},\ \ldots$$

$$a_n = (-1)^n: \quad -1,\ 1,\ -1,\ 1,\ \ldots$$

$$a_n = \frac{n+1}{n}: \quad 2,\ \frac{3}{2},\ \frac{4}{3},\ \frac{5}{4},\ \ldots$$

$$a_n = n^2: \quad 1,\ 4,\ 9,\ 16,\ \ldots$$

The central question about any sequence is: does it settle down? As $n$ gets larger and larger, do the terms $a_n$ approach some fixed value? Or do they wander off to infinity, oscillate forever, or behave chaotically?

This is the question of convergence.

Section 2: Convergence of Sequences

The sequence $a_n = 1/n$ starts at 1, goes to $1/2$, $1/3$, $1/4$, and the terms get smaller and smaller. They are approaching 0. We say the sequence converges to 0.

But what does “approach” mean, exactly? We need the same precision we built for limits of functions.

Definition. The sequence $(a_n)$ converges to the limit $L$ - written $\lim_{n \to \infty} a_n = L$ - if for every $\varepsilon > 0$, there exists a natural number $N$ such that for all $n > N$:

$$|a_n - L| < \varepsilon.$$

Unpack this. Your adversary names a tolerance $\varepsilon$: they demand that $a_n$ stay within $\varepsilon$ of $L$. You respond by finding a threshold $N$ beyond which every single term satisfies this demand. If you can always find such an $N$, no matter how small $\varepsilon$ is, the sequence converges to $L$.

If no such $L$ exists, the sequence diverges.

Example 1: $a_n = 1/n$ converges to 0.

Given $\varepsilon > 0$, choose $N$ to be any integer greater than $1/\varepsilon$. Then for $n > N$:

$$|a_n - 0| = \frac{1}{n} < \frac{1}{N} < \varepsilon.$$

Every term beyond the $N$-th is within $\varepsilon$ of 0. The sequence converges to 0.

Example 2: $a_n = (-1)^n$ diverges.

The terms are $-1, 1, -1, 1, \ldots$, alternating forever. Suppose some $L$ were the limit. Choose $\varepsilon = 1/2$. We need all terms eventually within $1/2$ of $L$. But the terms are always either $-1$ or $1$, which are distance 2 apart. They cannot both be within $1/2$ of any single number. No matter what $N$ we choose, among $a_N, a_{N+1}$ we find both $-1$ and $1$. The sequence diverges.

Example 3: $a_n = (n+1)/n$ converges to 1.

Write $a_n = 1 + 1/n$. As $n \to \infty$, the $1/n$ term vanishes and the sequence approaches 1. Formally: $|a_n - 1| = 1/n$, and by the same argument as Example 1, this can be made smaller than any $\varepsilon$.

Discomfort check. The definition says “there exists $N$ such that for all $n > N$…” The key word is “eventually.” The sequence does not need to be close to $L$ from the start. It can behave wildly for the first million terms and then settle down. Convergence is about the tail of the sequence, not its beginning. This matters when we work with series: the convergence of a series is entirely determined by its tail.

Section 3: The Monotone Convergence Theorem

Here is a situation that comes up constantly: a sequence is increasing, and you can prove it is bounded above, but you have no idea what it converges to. Maybe there is no formula for the limit.

Is that enough to guarantee convergence?

Yes. This is the Monotone Convergence Theorem, and it is deeper than it first appears.

Theorem. If a sequence $(a_n)$ is non-decreasing (each term is at least as large as the previous: $a_{n+1} \geq a_n$ for all $n$) and bounded above (there is some $M$ with $a_n \leq M$ for all $n$), then $(a_n)$ converges.

Furthermore, it converges to its supremum: $\lim_{n \to \infty} a_n = \sup\{a_n : n \geq 1\}$.

Why this is not obvious. If you have a bounded increasing sequence, where is it going? It cannot go above $M$. It keeps increasing. So it must be pressing up against some ceiling. But what if there is no ceiling it reaches - only a ceiling it approaches asymptotically? The theorem says that ceiling exists and is a real number. The sequence gets there.

Why it requires completeness. The theorem is actually a theorem about the real numbers. The statement that a bounded non-empty set has a supremum is the completeness of $\mathbb{R}$. On the rational numbers $\mathbb{Q}$, this fails. Consider the sequence of rational approximations to $\sqrt{2}$: $1, 1.4, 1.41, 1.414, \ldots$. This is increasing and bounded above (by 2, say). It satisfies the hypotheses of the theorem. But $\sqrt{2}$ is irrational - there is no rational number it converges to. The sequence diverges in $\mathbb{Q}$.

The real numbers were designed to fill precisely those gaps. Completeness - the property that bounded increasing sequences always have a limit - is not provable from simpler facts. It is baked into the construction of $\mathbb{R}$, and it is what makes analysis work.

The proof in one paragraph: let $L = \sup\{a_n\}$. This supremum exists by completeness. Given $\varepsilon > 0$, $L - \varepsilon$ is not an upper bound for the sequence (otherwise $L$ would not be the least upper bound), so there is some $a_N > L - \varepsilon$. Since the sequence is non-decreasing, for all $n > N$ we have $a_n \geq a_N > L - \varepsilon$. Also $a_n \leq L$ (it is bounded above by the supremum). So $|a_n - L| < \varepsilon$ for all $n > N$. $\blacksquare$

Discomfort check. “Prove convergence without knowing the limit” feels wrong. Usually when we prove something converges, we exhibit the limit. But sometimes you cannot - the limit exists as a real number but has no closed form. Euler’s number $e = \lim_{n \to \infty}(1 + 1/n)^n$ is exactly this situation. You can prove the sequence $(1 + 1/n)^n$ is increasing and bounded by 3 without knowing the limit is $e$. The Monotone Convergence Theorem guarantees the limit exists. Then we name it $e$. The theorem is powerful precisely because it decouples existence from identification.

Section 4: What Is a Series?

A series is an attempt to add up all the terms of a sequence. The series formed from $(a_n)$ is:

$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$$

But what does an infinite sum mean? We cannot actually add infinitely many numbers one by one. We define it through partial sums.

Definition. The $N$-th partial sum is the finite sum:

$$S_N = a_1 + a_2 + \cdots + a_N = \sum_{n=1}^{N} a_n.$$

The series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $(S_N)$ converges:

$$\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} S_N.$$

If $(S_N)$ diverges, the series diverges.

This is the bridge. The series is a question about a sequence - the sequence of partial sums. Everything we know about sequence convergence applies here.

Discomfort check. Why define series through partial sums and not just… add all the terms? Because “add all the terms” is not a mathematical operation unless you define it. The definition through partial sums is the definition. A series converges when the partial sums approach a finite limit. A series does not converge by some separate criterion and then we check the partial sums. The partial sums ARE the definition. Hold this distinction carefully - it will matter when we discuss rearrangements later.

Section 5: The Geometric Series - The Most Important Series

Let $a$ and $r$ be real numbers. The geometric series is:

$$a + ar + ar^2 + ar^3 + \cdots = \sum_{n=0}^{\infty} ar^n.$$

Every term is the previous term multiplied by the common ratio $r$.

When does it converge? We need to find the partial sums first.

$$S_N = a + ar + ar^2 + \cdots + ar^N.$$

Multiply both sides by $r$:

$$rS_N = ar + ar^2 + ar^3 + \cdots + ar^{N+1}.$$

Subtract:

$$S_N - rS_N = a - ar^{N+1}.$$

$$(1 - r)S_N = a(1 - r^{N+1}).$$

If $r \neq 1$:

$$S_N = \frac{a(1 - r^{N+1})}{1 - r}.$$

Now take $N \to \infty$. What happens to $r^{N+1}$?

  • If $|r| < 1$: $r^{N+1} \to 0$. So $S_N \to \frac{a}{1-r}$.
  • If $|r| > 1$: $|r^{N+1}| \to \infty$. The partial sums blow up. Diverges.
  • If $r = 1$: $S_N = (N+1)a \to \infty$ (assuming $a \neq 0$). Diverges.
  • If $r = -1$: terms alternate, $S_N$ oscillates. Diverges.

Theorem (Geometric Series). For $|r| < 1$:

$$\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}.$$

For $|r| \geq 1$, the series diverges.

Resolving Zeno’s paradox. The sum Zeno was computing is:

$$1 + \frac{1}{10} + \frac{1}{100} + \cdots = \sum_{n=0}^{\infty} \left(\frac{1}{10}\right)^n = \frac{1}{1 - 1/10} = \frac{10}{9}.$$

So Achilles runs $100 \cdot \frac{10}{9} = \frac{1000}{9} \approx 111.1$ meters total before catching the tortoise. An infinite number of steps, a finite total distance. The paradox was a confusion between “infinitely many steps” and “infinite total distance.” These are not the same thing. A geometric series with $|r| < 1$ is an infinite sum with a finite answer.

Another example. What is $0.\overline{9} = 0.999\ldots$?

$$0.9 + 0.09 + 0.009 + \cdots = \sum_{n=1}^{\infty} \frac{9}{10^n} = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1.$$

$0.\overline{9} = 1$. Exactly. Not approximately. This is a consequence of the geometric series formula. Many people find this deeply counterintuitive. The geometric series formula makes it airtight.

Section 6: The Harmonic Series - An Important Warning

Consider the series:

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots = \sum_{n=1}^{\infty} \frac{1}{n}.$$

This is called the harmonic series. The terms $1/n$ go to 0. You might think: the terms are getting smaller and smaller, so the sum must converge to something finite.

This is wrong. The harmonic series diverges.

This is one of the most important facts in the subject. The terms going to zero is necessary but not sufficient for convergence. Not nearly sufficient.

Proof (Oresme’s grouping argument, ca. 1350 CE): Group the terms as follows:

$$1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \left(\frac{1}{9} + \cdots + \frac{1}{16}\right) + \cdots$$

Look at each group:

  • Group 1: just $1$.
  • Group 2: just $1/2$.
  • Group 3: $1/3 + 1/4 > 1/4 + 1/4 = 1/2$.
  • Group 4: $1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 1/2$.
  • Group 5: the next 8 terms, each $\geq 1/16$, so sum $> 8 \cdot (1/16) = 1/2$.

Each group contributes more than $1/2$. There are infinitely many groups. The sum exceeds $1 + 1/2 + 1/2 + 1/2 + \cdots$, which diverges. So the harmonic series diverges. $\blacksquare$

This is deeply unintuitive. The terms $1/n$ are shrinking. After ten million terms, you are adding numbers smaller than $0.0000001$. But the sum still grows without bound - just incredibly slowly. By the time the partial sum first exceeds 10, you need more than $10^4$ terms. To exceed 100, you need more than $10^{43}$ terms. Divergence does not mean fast divergence. It just means unbounded.

Discomfort check. The harmonic series is the central cautionary example of this subject. “Terms go to zero” is a necessary condition for convergence - if the terms do not go to zero, the series definitely diverges. But it is not sufficient. The harmonic series proves this conclusively. Every time you want to argue a series converges because its terms are going to zero, stop. That argument does not work. You need a stronger test.

Section 7: The Divergence Test

The simplest test for divergence:

Theorem (Divergence Test). If $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n \to \infty} a_n = 0$.

Equivalently: if $a_n \not\to 0$, the series diverges.

Proof. Let $S = \lim_{N \to \infty} S_N$ be the sum. Then $a_N = S_N - S_{N-1} \to S - S = 0$. $\blacksquare$

When to use it. Use this test first, before anything else. It is the cheapest test.

  • $\sum \frac{n}{n+1}$: terms go to 1, not 0. Diverges immediately.
  • $\sum (-1)^n$: terms oscillate between $-1$ and $1$, do not go to 0. Diverges immediately.
  • $\sum \frac{1}{n}$: terms go to 0. Test is silent. Need a different tool.

The divergence test can only confirm divergence. It cannot confirm convergence.

Section 8: The Comparison Test

The idea: if every term of your series is smaller than the corresponding term of a series you know converges, your series converges. If every term is larger than a divergent series, it diverges.

Theorem (Comparison Test). Suppose $0 \leq a_n \leq b_n$ for all sufficiently large $n$.

  • If $\sum b_n$ converges, then $\sum a_n$ converges.
  • If $\sum a_n$ diverges, then $\sum b_n$ diverges.

When to use it. Use this when you can recognize that your series is termwise smaller than a geometric series, or termwise larger than the harmonic series.

Example. Does $\sum_{n=1}^{\infty} \frac{1}{n^2 + 5}$ converge?

Note $\frac{1}{n^2 + 5} \leq \frac{1}{n^2}$. If $\sum 1/n^2$ converges, so does our series. We know $\sum 1/n^2 = \pi^2/6$ (convergence follows from the integral test we will see shortly). So yes, $\sum \frac{1}{n^2+5}$ converges.

Example. Does $\sum_{n=2}^{\infty} \frac{1}{\ln n}$ converge?

Note $\frac{1}{\ln n} \geq \frac{1}{n}$ for $n \geq 3$ (since $\ln n \leq n$). Since $\sum 1/n$ diverges and $1/\ln n \geq 1/n$, the comparison test says $\sum 1/\ln n$ diverges.

Section 9: The Ratio Test

The comparison test is powerful but requires finding a comparison series. The ratio test is more automatic: it looks at the ratio of consecutive terms.

Theorem (Ratio Test). Let $a_n > 0$ for all $n$, and suppose the limit

$$\rho = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$$

exists.

  • If $\rho < 1$: the series $\sum a_n$ converges.
  • If $\rho > 1$: the series $\sum a_n$ diverges.
  • If $\rho = 1$: the test is inconclusive.

Why does this work? If $\rho < 1$, pick $r$ with $\rho < r < 1$. Eventually, $a_{n+1}/a_n < r$ for all large $n$. This means the terms are shrinking at least as fast as a geometric series with ratio $r$. More precisely: for $n \geq N$,

$$a_{N+k} \leq a_N \cdot r^k.$$

Summing over $k$:

$$\sum_{k=0}^{\infty} a_{N+k} \leq a_N \sum_{k=0}^{\infty} r^k = \frac{a_N}{1-r} < \infty.$$

The tail converges. Since the tail converges, the whole series converges.

If $\rho > 1$, eventually $a_{n+1} > a_n$, so the terms are growing. They cannot go to zero. By the divergence test, the series diverges.

The ratio test detects geometric-like behavior. A series converges by the ratio test when its terms, in ratio, look like a geometric series with ratio less than 1. This is why the ratio test is especially effective for series involving factorials and exponentials - things that naturally grow or shrink by a constant multiplicative factor each time.

Example. Does $\sum_{n=0}^{\infty} \frac{n!}{3^n}$ converge?

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)! / 3^{n+1}}{n! / 3^n} = \frac{(n+1)}{3} \to \infty.$$

$\rho = \infty > 1$. Diverges.

Example. Does $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ converge for any fixed $x$?

$$\frac{a_{n+1}}{a_n} = \frac{|x|^{n+1}/(n+1)!}{|x|^n/n!} = \frac{|x|}{n+1} \to 0.$$

$\rho = 0 < 1$. Converges for every $x$. This is the exponential series $e^x$ - it converges everywhere on the real line.

The case $\rho = 1$. When the ratio test gives $\rho = 1$, it tells you nothing. Both $\sum 1/n$ (divergent) and $\sum 1/n^2$ (convergent) give $\rho = 1$ in the ratio test. The test is blind to power-series decay.

Section 10: The Integral Test

For series whose terms are values of a decreasing function, the series and a related integral converge or diverge together.

Theorem (Integral Test). Let $f: [1, \infty) \to \mathbb{R}$ be a continuous, positive, decreasing function. Define $a_n = f(n)$. Then:

$$\sum_{n=1}^{\infty} a_n \text{ converges} \iff \int_1^{\infty} f(x)dx \text{ converges}.$$

Why? Because $f$ is decreasing, on the interval $[n, n+1]$, the function satisfies $f(n+1) \leq f(x) \leq f(n)$. Integrating over $[n, n+1]$:

$$f(n+1) \leq \int_n^{n+1} f(x)dx \leq f(n).$$

Sum from $n = 1$ to $N-1$:

$$\sum_{n=2}^{N} f(n) \leq \int_1^N f(x)dx \leq \sum_{n=1}^{N-1} f(n).$$

The partial sums and the integral are sandwiched between each other. If one is bounded, so is the other. They converge or diverge together.

The p-series. The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$.

Proof: Apply the integral test. $\int_1^{\infty} \frac{1}{x^p} dx$ converges if and only if $p > 1$ (this is a standard improper integral computation).

So:

  • $\sum 1/n^2 = \pi^2/6$. Converges.
  • $\sum 1/n^{3/2}$. Converges.
  • $\sum 1/n$. Diverges ($p = 1$).
  • $\sum 1/\sqrt{n}$. Diverges ($p = 1/2 < 1$).

The harmonic series diverges because $p = 1$ is exactly the borderline case.

Section 11: The Alternating Series Test

All tests so far apply to series with non-negative terms. But many important series have mixed signs.

Theorem (Alternating Series Test, Leibniz). Suppose $(b_n)$ is a sequence of positive numbers that is decreasing ($b_{n+1} \leq b_n$ for all $n$) and converging to zero ($b_n \to 0$). Then the series:

$$\sum_{n=1}^{\infty} (-1)^{n+1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots$$

converges.

Why? Look at the even partial sums $S_{2k}$. Each pair of terms contributes:

$$b_{2k-1} - b_{2k} = b_{2k-1} - b_{2k} \geq 0$$

since $b_{2k-1} \geq b_{2k}$ (the sequence is decreasing). So $(S_{2k})$ is non-decreasing. Also, $S_{2k} = b_1 - (b_2 - b_3) - (b_4 - b_5) - \cdots \leq b_1$ (each parenthesized term is non-negative). Bounded and non-decreasing: by the Monotone Convergence Theorem, $(S_{2k})$ converges to some $L$.

The odd partial sums: $S_{2k+1} = S_{2k} + b_{2k+1} \to L + 0 = L$ since $b_{2k+1} \to 0$.

Both subsequences converge to $L$, so the whole sequence of partial sums converges to $L$. $\blacksquare$

The error bound. If you stop at the $N$-th term, the error is at most $b_{N+1}$:

$$\left|\sum_{n=1}^{\infty} (-1)^{n+1} b_n - S_N\right| \leq b_{N+1}.$$

This gives alternating series excellent numerical properties. Stopping at the first term you drop gives an error bounded by that term.

Example. The alternating harmonic series:

$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}.$$

The $b_n = 1/n$ are decreasing and go to zero. The alternating series test guarantees convergence. The sum is $\ln 2 \approx 0.693$.

Section 12: Absolute and Conditional Convergence

The alternating harmonic series converges. The ordinary harmonic series diverges. What should we make of a series that converges, but only because of cancellation between positive and negative terms?

Definition. The series $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges.

Definition. The series $\sum a_n$ is conditionally convergent if it converges but $\sum |a_n|$ diverges.

Theorem. Absolute convergence implies convergence.

Proof. If $\sum |a_n|$ converges, then for any $n$: $0 \leq a_n + |a_n| \leq 2|a_n|$. By comparison, $\sum (a_n + |a_n|)$ converges. Then $\sum a_n = \sum(a_n + |a_n|) - \sum |a_n|$ is the difference of two convergent series, hence convergent. $\blacksquare$

The alternating harmonic series is conditionally convergent: it converges, but its absolute value series (the harmonic series) diverges.

Now for the deeply disturbing fact.

Theorem (Riemann Rearrangement Theorem). If $\sum a_n$ is conditionally convergent, then for any real number $S$ (including $\pm \infty$), there is a rearrangement of the terms of $\sum a_n$ that sums to $S$.

A rearrangement is a permutation of the terms. You take all the same terms but add them in a different order.

This means the alternating harmonic series, which sums to $\ln 2 \approx 0.693$, can be rearranged to sum to 47. Or to $-\pi$. Or to any number you like.

This is not a parlor trick. It is a genuine mathematical fact, and it should disturb you.

How the proof works. Separate the positive and negative terms. The positive terms of $\sum a_n$ form a series $\sum p_n$ and the negative terms form a series $\sum q_n$. If $\sum a_n$ is conditionally convergent, then $\sum p_n = +\infty$ and $\sum q_n = -\infty$ (both diverge). This is the key.

To rearrange to sum to $S > 0$: take positive terms until the partial sum exceeds $S$. Then take negative terms until it falls below $S$. Then positive terms until it exceeds $S$ again. Alternate. Since both $\sum p_n$ and $\sum q_n$ diverge, we never run out of terms from either pool. The oscillations get smaller (since the individual terms go to zero), so the partial sums converge to $S$.

The moral. For conditionally convergent series, the sum depends on the order in which you add the terms. Addition of infinitely many numbers is not always commutative. This is deeply contrary to intuition - for finite sums, you can add in any order. For infinite sums, rearrangement can change everything.

For absolutely convergent series, there is no such danger. Absolutely convergent series can be rearranged freely without changing the sum. This is why absolute convergence is the “safe” kind of convergence.

Discomfort check. The Riemann rearrangement theorem is one of those results that you should sit with until it genuinely bothers you. The alternating harmonic series $1 - 1/2 + 1/3 - 1/4 + \cdots$ converges to $\ln 2$. Write the same terms in a different order and you get $\pi$. Or $-100$. Or $0$. The sum is not a property of the terms. It is a property of the terms together with their order. This is a fundamental reason why, in analysis, we care enormously about the distinction between absolute and conditional convergence.

Section 13: Power Series

So far we have been summing numbers. Now let $x$ be a variable. A power series centered at $c$ is a series of the form:

$$\sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x - c) + a_2(x - c)^2 + a_3(x - c)^3 + \cdots$$

For a fixed value of $x$, this is just a number series. It either converges or diverges. But $x$ is free to vary, so we can ask: for which values of $x$ does this series converge?

The Radius of Convergence. The answer always has a beautiful structure: there is a number $R \geq 0$ (possibly $\infty$) such that:

  • The series converges absolutely for all $x$ with $|x - c| < R$.
  • The series diverges for all $x$ with $|x - c| > R$.
  • At the endpoints $|x - c| = R$, behavior must be checked separately.

The number $R$ is called the radius of convergence. The interval $(c - R, c + R)$ is the interval of convergence (possibly extended at the endpoints).

Finding $R$ via the ratio test. If you apply the ratio test to the power series, looking at $|a_{n+1}(x-c)^{n+1}| / |a_n(x-c)^n| = |a_{n+1}/a_n| \cdot |x-c|$, the test gives convergence when this limit is less than 1:

$$\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} \cdot |x - c| < 1.$$

This gives $|x - c| < \frac{1}{\lim |a_{n+1}/a_n|}$, so $R = \lim_{n \to \infty} \frac{|a_n|}{|a_{n+1}|}$ (or equivalently $1 / \limsup |a_n|^{1/n}$ by the root test - the Cauchy-Hadamard formula).

Examples.

The series $\sum_{n=0}^{\infty} x^n$ is geometric with ratio $x$. It converges for $|x| < 1$. Radius of convergence $R = 1$.

The series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ converges for all $x$ (we showed this in the ratio test section). Radius of convergence $R = \infty$.

The series $\sum_{n=0}^{\infty} n! \cdot x^n$ has ratio $|(n+1)! x^{n+1}| / |n! x^n| = (n+1)|x| \to \infty$ for any $x \neq 0$. Converges only at $x = 0$. Radius of convergence $R = 0$.

Power series define functions. On its interval of convergence, a power series defines a function $f(x) = \sum a_n (x-c)^n$. This function is infinitely differentiable, and we can differentiate or integrate term by term:

$$f'(x) = \sum_{n=1}^{\infty} n \cdot a_n (x-c)^{n-1},$$

$$\int f(x)dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1} (x-c)^{n+1}.$$

Both of these have the same radius of convergence $R$. Differentiation and integration of power series work term by term - just like finite polynomials.

This is the bridge to Taylor series. If a function $f$ can be represented as a power series $\sum a_n (x-c)^n$, then we can determine the coefficients by differentiating:

$$f^{(k)}(c) = k! \cdot a_k \implies a_k = \frac{f^{(k)}(c)}{k!}.$$

The coefficients are determined by the derivatives at the center. That observation is the entire idea of Taylor series - which is the next post.

Section 14: Why Convergence Is Hard - A Final Note

We now have six tests: divergence, comparison, ratio, integral, alternating series, and the radius of convergence formula for power series. This feels like a lot. And yet: there are series that converge, but for which none of these tests applies cleanly.

The theory of series is genuinely hard. The reason is that convergence is about the behavior of partial sums as $N \to \infty$ - a global question about an infinite process. Local information about individual terms (how fast $a_n$ goes to zero, whether $a_n$ is positive) is often not enough.

The tests we have are tools for common patterns. Learn when each applies:

  • Divergence test: first. If terms do not go to zero, stop.
  • Ratio test: when you have factorials, exponentials, or terms that have a clean ratio.
  • Comparison: when you can bound your series above or below by something you know.
  • Integral test: when the terms come from a decreasing function and an integral is easier to compute.
  • Alternating series test: when terms alternate and decrease to zero.
  • Absolute convergence: when you can show $\sum |a_n|$ converges, you are done.

The harmonic series sits as a warning behind all of them. Fast-looking decay can still diverge. Slow-looking decay can still converge. The subject rewards careful analysis and punishes intuition.

Summary

Concept Definition Key fact
Sequence convergence $\lim_{n \to \infty} a_n = L$ means terms eventually stay within $\varepsilon$ of $L$ Increasing and bounded $\Rightarrow$ converges (MCT)
Series convergence $\sum a_n$ converges if partial sums $S_N \to L$ Defined entirely through partial sums
Geometric series $\sum ar^n = a/(1-r)$ for $|r| < 1$ The fundamental computable series; resolves Zeno
Harmonic series $\sum 1/n$ diverges Terms $\to 0$ is necessary but not sufficient
p-series $\sum 1/n^p$ converges iff $p > 1$ Boundary case $p=1$ diverges
Divergence test $a_n \not\to 0 \Rightarrow$ diverges Cannot confirm convergence
Ratio test $\rho = \lim |a_{n+1}/a_n|$; converges if $\rho < 1$ Best for factorials and exponentials
Integral test Series and integral converge together Best for $p$-series and similar
Alternating series Decreasing terms $\to 0 \Rightarrow$ converges Error $\leq$ first omitted term
Absolute convergence $\sum |a_n| < \infty \Rightarrow \sum a_n$ converges Rearrangement-safe
Conditional convergence Converges but not absolutely Riemann: can rearrange to any sum
Power series $\sum a_n(x-c)^n$, converges on $|x - c| < R$ Foundation for Taylor series

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