Infinite Sequences and Series

What does it mean to add up infinitely many numbers and get a finite answer? That single question — half sane, half paradoxical — is the doorway into one of the most powerful ideas in all of calculus. When you write $0.999\ldots = 1$, or when your calculator computes $\sin(0.7)$ or $e^{2}$, you are quietly relying on infinite series. They let us build complicated functions out of nothing but additions and multiplications, and they let us tame infinity with precise, checkable rules.

The catch is that infinity does not play fair. Some infinite sums settle down to a number (they converge); others grow without bound or oscillate forever (they diverge). Learning this topic means learning to tell the two apart with confidence, and then learning to use the convergent ones to represent functions. This is the machinery behind polynomial approximation, numerical computing, physics, and much of higher mathematics.

Learning Objectives

• Distinguish a sequence from a series, and define convergence for each precisely.
• Compute sums of geometric series and identify when they converge.
• Apply the core convergence tests: $n$th-term (divergence) test, geometric, $p$-series, comparison, limit comparison, ratio, and integral tests.
• Determine the radius and interval of convergence of a power series.
• Construct Taylor and Maclaurin series and use them to approximate functions.
• Appreciate why rigor — Cauchy's definition of a limit — was historically necessary.

Quick Answer

A sequence is an ordered list of numbers $a_1, a_2, a_3, \ldots$; a series is what you get by adding a sequence's terms, $\sum a_n$. A series converges if its partial sums $S_N = a_1 + \cdots + a_N$ approach a finite limit as $N \to \infty$. The geometric series $\sum ar^{n}$ converges to $\frac{a}{1-r}$ exactly when $|r| < 1$. The $p$-series $\sum \frac{1}{n^{p}}$ converges when $p > 1$. To test a general series, choose a tool: the $n$th-term test rules out obvious divergence, while the comparison, ratio, and integral tests handle the subtler cases. Finally, a power series $\sum c_n (x-a)^{n}$ turns a function into an "infinite polynomial," and its most important instance is the Taylor series, which reconstructs a function from its derivatives at a single point.

Where It Came From

The paradox of infinite sums is ancient. Around 450 BC, Zeno of Elea argued that motion is impossible: to cross a room you must first cross half, then half of what remains, then half again — infinitely many steps that (he claimed) could never be completed. Buried in that puzzle is the geometric series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1$. The Greeks lacked the language to say that infinitely many positive terms could sum to something finite, so the paradox stood for two millennia.

The real motivation for developing series arrived with the scientific revolution. In the 1660s, Isaac Newton needed to compute areas, roots, and values of functions like $\sqrt{1+x}$ and $\ln(1+x)$ for which no simple formula existed. His breakthrough was to treat these functions as infinite polynomials — expand them into series and integrate or evaluate term by term. This made previously intractable calculations routine and became a foundation of his calculus.

A generation later, Leonhard Euler (1700s) turned series into an art form. He summed the notorious Basel problem, proving $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$, and discovered the stunning identity $e^{i\theta} = \cos\theta + i\sin\theta$ by manipulating power series. Euler's results were correct and dazzling — but his methods were often bold to the point of recklessness, treating divergent series as if they had ordinary sums.

That looseness eventually caused contradictions, and the 1800s brought a demand for rigor. Augustin-Louis Cauchy gave the modern definition of convergence in terms of partial sums and $\varepsilon$-precision limits, cleanly separating series that genuinely converge from those that merely look tidy. His work — later sharpened by Weierstrass — is why we now test convergence instead of assuming it. The lesson of this history is baked into the topic: infinite processes reward you enormously, but only if you check the rules first.

Sequences, Series, and What Convergence Means

A sequence is a function whose domain is the positive integers: $a_n = \frac{1}{n}$ gives $1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots$. It converges to $L$ if the terms get arbitrarily close to $L$; here $a_n \to 0$.

A series sums a sequence. The key idea is the sequence of partial sums:

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

The infinite series $\sum a_n$ is said to converge to $S$ if $\lim_{N \to \infty} S_N = S$. Convergence of a series is defined entirely through the sequence of its partial sums — this is Cauchy's crucial move.

The $n$th-term test (divergence test). If $\lim_{n\to\infty} a_n \neq 0$, the series $\sum a_n$ must diverge. Reason: for a sum to settle, the pieces being added must shrink to nothing. Warning — this is a one-way test. If $a_n \to 0$, the series might converge or might not; the test says nothing.

Worked example. Does $\sum_{n=1}^{\infty} \frac{n}{2n+1}$ converge? Look at the terms: $\lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2} \neq 0$. Since the terms do not go to zero, the series diverges. Done — no deeper test needed.

The Two Series You Must Know Cold

Geometric series. $\displaystyle\sum_{n=0}^{\infty} a r^{n} = a + ar + ar^{2} + \cdots$. Using the identity $S_N = \frac{a(1-r^{N+1})}{1-r}$, we see that if $|r| < 1$ then $r^{N+1} \to 0$ and

$\sum_{n=0}^{\infty} a r^{n} = \frac{a}{1-r}, \qquad |r| < 1.$

If $|r| \ge 1$ the terms do not shrink to zero, so it diverges.

Worked example. Zeno, resolved: $\sum_{n=1}^{\infty} \left(\tfrac{1}{2}\right)^{n} = \tfrac{1}{2} + \tfrac{1}{4} + \cdots$. Here first term $a = \tfrac{1}{2}$ and ratio $r = \tfrac{1}{2}$, so the sum is $\frac{1/2}{1 - 1/2} = 1$. You do finish crossing the room.

$p$-series. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{p}}$ converges if and only if $p > 1$. Two landmark cases: the harmonic series $\sum \frac{1}{n}$ ($p = 1$) diverges even though its terms go to zero — the classic trap — while $\sum \frac{1}{n^{2}}$ ($p = 2$) converges (to $\frac{\pi^{2}}{6}$, Euler's result).

Choosing and Using a Convergence Test

For series that aren't obviously geometric or $p$-series, pick the right tool.

Integral test. If $f(x)$ is positive, continuous, and decreasing with $f(n) = a_n$, then $\sum a_n$ and $\int_1^{\infty} f(x)\,dx$ converge or diverge together.

Worked example. Test $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$. Let $f(x) = \frac{1}{x\ln x}$. Then

$\int_{2}^{\infty} \frac{1}{x \ln x}\,dx = \Big[\ln(\ln x)\Big]_{2}^{\infty} = \infty.$

The integral diverges, so the series diverges.

Comparison and limit comparison tests. If $0 \le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges (a smaller positive series can't misbehave under a well-behaved one). The limit comparison test is often easier: if $\lim_{n\to\infty} \frac{a_n}{b_n} = c$ with $0 < c < \infty$, then $\sum a_n$ and $\sum b_n$ share the same fate.

Worked example. Test $\sum \frac{1}{n^{2}+3n}$. Compare with $b_n = \frac{1}{n^{2}}$. Then $\frac{a_n}{b_n} = \frac{n^{2}}{n^{2}+3n} = \frac{1}{1 + 3/n} \to 1$. Since $\sum \frac{1}{n^{2}}$ converges ($p = 2 > 1$), the original series converges.

Ratio test. Compute $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$. If $L < 1$ the series converges (absolutely); if $L > 1$ it diverges; if $L = 1$ the test is inconclusive. This is the workhorse for factorials and exponentials.

Worked example. Test $\sum_{n=1}^{\infty} \frac{2^{n}}{n!}$.

$L = \lim_{n\to\infty} \frac{2^{n+1}/(n+1)!}{2^{n}/n!} = \lim_{n\to\infty} \frac{2}{n+1} = 0 < 1.$

So the series converges (in fact its sum is $e^{2} - 1$).

Power Series, Taylor, and Maclaurin

A power series centered at $a$ is $\displaystyle\sum_{n=0}^{\infty} c_n (x-a)^{n}$. For each $x$ it is just a number-series, so it converges for some $x$ and diverges for others. The set of $x$ where it converges is an interval centered at $a$, described by the radius of convergence $R$. The ratio test usually finds $R$.

Worked example (radius of convergence). For $\sum_{n=0}^{\infty} \frac{x^{n}}{n!}$, apply the ratio test to the terms:

$\lim_{n\to\infty} \left|\frac{x^{n+1}/(n+1)!}{x^{n}/n!}\right| = \lim_{n\to\infty} \frac{|x|}{n+1} = 0 < 1 \ \text{for all } x.$

So $R = \infty$: it converges everywhere. (It equals $e^{x}$.)

Taylor series. The deepest use of power series is representing a known function. If $f$ has derivatives of all orders at $a$, its Taylor series is

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^{n}.$

When $a = 0$ this is called a Maclaurin series. The intuition: match the function's value, slope, curvature, and every higher derivative at one point, and the resulting infinite polynomial reproduces the whole function (within its radius of convergence).

Worked example. Build the Maclaurin series of $e^{x}$. Every derivative of $e^{x}$ is $e^{x}$, and $e^{0} = 1$, so $f^{(n)}(0) = 1$ for all $n$:

$e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \cdots$

To estimate $e^{0.1}$, take the first three terms: $1 + 0.1 + \frac{0.01}{2} = 1.105$. The true value is $1.10517\ldots$ — accurate to four decimals with almost no work. Two other essentials worth memorizing:

$\sin x = x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots, \qquad \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n}\ (|x|<1).$

Real-World Applications

• Calculators and computers evaluate $\sin$, $\cos$, $e^{x}$, and $\ln$ using truncated Taylor series (or close relatives) — polynomials are all a CPU can really compute.
• Physics relies on approximations like $\sin\theta \approx \theta$ (the first Taylor term), which linearizes the pendulum equation and makes small-oscillation analysis solvable.
• Engineering signal processing decomposes waveforms into series (Fourier series), a direct descendant of the idea that functions can be built from infinite sums.
• Finance uses geometric series for the present value of perpetuities and annuities: a payment stream discounted at rate $r$ sums geometrically.
• Numerical error control: the remainder term of a Taylor series tells engineers exactly how many terms guarantee a required precision.

Common Mistakes

1. Believing $a_n \to 0$ guarantees convergence. Why it's wrong: The harmonic series $\sum \frac{1}{n}$ has terms shrinking to zero yet diverges to infinity. Correction: $a_n \to 0$ is necessary but not sufficient. Use it only to prove divergence (when the limit isn't zero), never to conclude convergence.

2. Applying the geometric formula when $|r| \ge 1$. Why it's wrong: $\frac{a}{1-r}$ is derived assuming $r^{N+1} \to 0$, which fails when $|r| \ge 1$. Writing $\sum 2^{n} = \frac{1}{1-2} = -1$ is nonsense — the sum is infinite. Correction: Always check $|r| < 1$ before using the sum formula.

3. Forgetting to check the endpoints of a power series' interval. Why it's wrong: The ratio test gives the radius $R$ but is inconclusive at $x = a \pm R$. The series may converge at one endpoint, both, or neither. Correction: After finding $R$, substitute each endpoint separately and test those number-series directly.

Comparison and Connections

Sequences and series are constantly confused, and so are the tests. This table keeps them straight.

Concept	What it is	Converges when
Sequence $a_n$	Ordered list of terms	Terms approach a single limit $L$
Series $\sum a_n$	Sum of the terms	Partial sums $S_N$ approach a limit
Geometric $\sum ar^{n}$	Constant ratio $r$	$\vert r\vert < 1$
$p$-series $\sum n^{-p}$	Power of $n$ in denominator	$p > 1$
Ratio test	Uses $\vert a_{n+1}/a_n\vert$	Limit $L < 1$

A subtle connection: absolute vs. conditional convergence. If $\sum |a_n|$ converges, so does $\sum a_n$ (absolute convergence). But some series like $\sum \frac{(-1)^{n+1}}{n} = \ln 2$ converge only because of sign cancellation — the absolute version diverges. These are conditionally convergent, and reordering their terms can change the sum (Riemann's rearrangement theorem).

Practice Questions

Recall

State the condition under which the geometric series $\sum_{n=0}^{\infty} ar^{n}$ converges, and give its sum. Answer: It converges iff $|r| < 1$, with sum $\frac{a}{1-r}$.

Understanding

Explain why the harmonic series $\sum \frac{1}{n}$ diverges even though $\frac{1}{n} \to 0$. Guidance: The $n$th-term test only rules out convergence when terms don't vanish; vanishing terms prove nothing. Group terms: $\frac{1}{3}+\frac{1}{4} > \frac{1}{2}$, $\frac{1}{5}+\cdots+\frac{1}{8} > \frac{1}{2}$, and so on — infinitely many blocks each exceeding $\frac{1}{2}$, so the partial sums grow without bound.

Application

Find the radius of convergence of $\sum_{n=1}^{\infty} \frac{(x-2)^{n}}{n \, 3^{n}}$. Answer: Ratio test: $\left|\frac{(x-2)^{n+1}/((n+1)3^{n+1})}{(x-2)^{n}/(n 3^{n})}\right| = \frac{n}{n+1}\cdot\frac{|x-2|}{3} \to \frac{|x-2|}{3}$. Converges when $\frac{|x-2|}{3} < 1$, so $R = 3$ (centered at $x = 2$).

Analysis

Determine whether $\sum_{n=1}^{\infty} \frac{n!}{n^{n}}$ converges. Answer: Ratio test: $\frac{(n+1)!/(n+1)^{n+1}}{n!/n^{n}} = \frac{(n+1) n^{n}}{(n+1)^{n+1}} = \left(\frac{n}{n+1}\right)^{n} = \frac{1}{(1+1/n)^{n}} \to \frac{1}{e} < 1$. It converges.

FAQ

Is $0.999\ldots$ really equal to 1? Yes, exactly. It is the geometric series $\frac{9}{10} + \frac{9}{100} + \cdots$ with $a = \frac{9}{10}$, $r = \frac{1}{10}$, summing to $\frac{9/10}{1 - 1/10} = 1$. Not "almost" — equal.

How do I choose which test to use? Scan the terms first. Terms not going to zero? Use the $n$th-term test. Factorials or $n$th powers? Ratio test. Looks like a $p$-series or a rational function of $n$? Comparison or limit comparison. Involves a function you can integrate? Integral test. Geometric shape? Geometric formula.

What's the difference between a Taylor and a Maclaurin series? None in kind — a Maclaurin series is just a Taylor series centered at $a = 0$. Center elsewhere when you want accuracy near a different point.

Why does the interval of convergence matter? A power series only equals the function it represents inside its interval. Using $\frac{1}{1-x} = \sum x^{n}$ at $x = 2$ gives garbage, because $|x| < 1$ fails there.

Can rearranging the terms of a series change its sum? For absolutely convergent series, no — order never matters. For conditionally convergent ones, yes: Riemann proved you can rearrange them to sum to any value you like. This is exactly why Cauchy's rigor was needed.

Quick Revision

• Series converges $\iff$ partial sums $S_N$ have a finite limit.
• $n$th-term test: $a_n \not\to 0 \Rightarrow$ diverges (but $a_n \to 0$ proves nothing).
• Geometric: converges iff $|r| < 1$; sum $= \frac{a}{1-r}$.
• $p$-series: $\sum n^{-p}$ converges iff $p > 1$; harmonic ($p=1$) diverges.
• Ratio test: $L < 1$ converge, $L > 1$ diverge, $L = 1$ inconclusive.
• Integral test: series and $\int f$ agree, for positive decreasing $f$.
• Taylor: $f(x) = \sum \frac{f^{(n)}(a)}{n!}(x-a)^{n}$; Maclaurin is $a = 0$.
• Memorize: $e^{x} = \sum \frac{x^{n}}{n!}$, $\frac{1}{1-x} = \sum x^{n}$.

Related Topics

Prerequisites

• Limits — convergence is defined through limits.
• Derivatives — Taylor coefficients are built from derivatives.

Related Topics

• Integrals — the integral test and term-by-term integration.
• Calculus overview — where series fit in the broader subject.

Next Topics

• Fourier series and advanced approximation (building functions from infinite sums of waves).
• Differential equations solved via power series methods.