Infinite Series & Taylor Polynomials

1. Introduction

Can you add up infinitely many numbers and get a finite total? Sometimes yes, sometimes no — and the entire theory of infinite series is about telling the two cases apart and, when possible, computing the sum. This question underlies how calculators evaluate $\sin$, $e^x$, and $\ln$, how engineers approximate complicated functions by polynomials, and how physicists expand models in small parameters.

The payoff is the Taylor series, which represents a function as an infinite polynomial built from its derivatives at a single point. Truncating yields Taylor polynomials — the best polynomial approximations near a point — with a remainder term quantifying the error. To get there we need sequences and series, benchmark examples (geometric and $p$-series), and convergence tests: divergence test, integral test, comparison and limit comparison, alternating series test, and the ratio and root tests.

The convergence landscape has a clear hierarchy: first check whether $a_n \to 0$ (divergence test); then try to recognize geometric or $p$-series; for factorials and exponentials use the ratio test; for $n$th powers use the root test; for positive terms "close to" a benchmark use limit comparison; for alternating signs use the alternating series test. Power series add the step of finding the radius of convergence and testing endpoints separately — the ratio test gives $|x-a| < R$ but is silent at $|x-a| = R$.

2. Core Concepts

2.1 Sequences vs. Series

A sequence $\{a_n\}$ converges if $\lim_{n\to\infty} a_n$ exists. A series $\sum_{n=1}^{\infty} a_n$ is the sum of terms, defined as the limit of partial sums $S_N = \sum_{n=1}^{N} a_n$. The series converges to $S$ if $\lim_{N\to\infty} S_N = S$; otherwise it diverges.

2.2 The Divergence Test

If $\lim_{n\to\infty} a_n \neq 0$ (or fails to exist), then $\sum a_n$ diverges.

Proof sketch: If $\sum a_n$ converges to $S$, then $S_N \to S$ and $S_{N-1} \to S$, so $a_N = S_N - S_{N-1} \to 0$. Contrapositive: if $a_n \not\to 0$, the series diverges.

This is necessary, not sufficient: $\sum \frac{1}{n}$ has $a_n \to 0$ but diverges.

2.3 Geometric Series

$\sum_{n=0}^{\infty} ar^n$ converges iff $|r| < 1$, with sum $\sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r}.$

Proof sketch: $S_N = a\frac{1-r^{N+1}}{1-r}$. If $|r| < 1$, $r^{N+1} \to 0$, so $S_N \to \frac{a}{1-r}$.

2.4 The $p$-Series and the Integral Test

$\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges iff $p > 1$.

Integral Test: If $f$ is positive, continuous, and decreasing on $[1,\infty)$ with $f(n) = a_n$, then $\sum_{n=1}^{\infty} a_n$ and $\int_1^{\infty} f(x)\,dx$ both converge or both diverge.

Proof sketch ($p$-series): $\int_1^{\infty} x^{-p}\,dx$ converges iff $p > 1$ (gives $\frac{1}{p-1}$), diverges iff $p \le 1$. By the integral test, so does $\sum 1/n^p$.

2.5 Comparison Tests

Direct Comparison: If $0 \le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges. If $a_n \ge b_n \ge 0$ and $\sum b_n$ diverges, so does $\sum a_n$.

Limit Comparison: If $a_n, b_n > 0$ and $\lim_{n\to\infty}\frac{a_n}{b_n} = L$ with $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ share the same fate.

Proof sketch (limit comparison): For large $n$, $a_n \approx L b_n$; the partial sums of $\sum a_n$ are roughly $L$ times those of $\sum b_n$.

2.6 Alternating Series Test

For $\sum (-1)^{n-1} b_n$ with $b_n > 0$, if $b_n$ is decreasing and $b_n \to 0$, the series converges. The error after $N$ terms satisfies $|S - S_N| \le b_{N+1}$.

Proof sketch: Partial sums $S_{2m}$ are increasing and bounded above (telescoping pairing); $S_{2m+1}$ decreasing and bounded below; both converge to the same limit.

2.7 Absolute and Conditional Convergence

Absolutely convergent if $\sum |a_n|$ converges (implies convergence). Conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges (e.g. alternating harmonic series). Rearranging a conditionally convergent series can change its sum.

2.8 The Ratio Test

Let $L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$. If $L < 1$, the series converges absolutely; if $L > 1$, it diverges; if $L = 1$, inconclusive.

Proof sketch (idea): If $L < 1$, pick $r$ with $L < r < 1$; for large $n$, $|a_{n+1}| < r|a_n|$, so terms decay like a geometric series.

2.9 The Root Test

Let $L = \lim_{n\to\infty}\sqrt[n]{|a_n|}$. Same conclusions as ratio test: $L < 1$ converges, $L > 1$ diverges, $L = 1$ inconclusive.

2.10 Power Series and Radius of Convergence

$\sum_{n=0}^{\infty} c_n (x-a)^n$ converges on an interval centered at $a$. The radius of convergence $R$ is found by the ratio test: solve $\lim\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| < 1$. Inside $(a-R, a+R)$ the sum defines an analytic function.

2.11 Taylor Series and Remainder

The Taylor series of $f$ about $a$ is $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n.$ Taylor's Theorem (Lagrange remainder): if $f^{(N+1)}$ is bounded by $M$ on the interval between $a$ and $x$, then $|R_N(x)| = |f(x) - T_N(x)| \le \frac{M}{(N+1)!}|x-a|^{N+1}.$

Proof sketch (remainder): Apply the MVT or integration by parts repeatedly to express $R_N$ as an integral of $f^{(N+1)}$, then bound the integral.

2.12 Operations on Power Series

Inside the interval of convergence, power series can be differentiated and integrated term-by-term: $\frac{d}{dx}\sum c_n x^n = \sum n c_n x^{n-1}, \qquad \int \sum c_n x^n\,dx = C + \sum \frac{c_n}{n+1}x^{n+1}.$ The radius of convergence is unchanged by differentiation or integration (endpoints may change).

2.13 Taylor Series from Known Expansions

Rather than computing derivatives, derive new series by substitution: replace $x$ with $-x^2$ in $\frac{1}{1-x}$ to get $\frac{1}{1+x^2} = \sum (-1)^n x^{2n}$. Integrate term-by-term to recover $\arctan x$. Differentiate $\sum x^n$ to get $\sum n x^{n-1}$.

2.14 Cauchy Condensation Test (Preview)

For positive decreasing $a_n$, $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges. This gives an elegant proof that $\sum \frac{1}{n\ln n}$ diverges.