Infinite Sequences and Series: From Convergence to Taylor Series
Published:
By: Rajib Belbase
For students in Calculus II. This post covers the essential ideas behind infinite sequences and series — what they mean, how to test for convergence, and how to represent functions as power series. All material follows Chapter 11 of Calculus, 9th edition, by James Stewart.
1. Infinite Sequences
An infinite sequence is an ordered list of numbers, one for each positive integer:
\[\{a_n\}_{n=1}^{\infty} = \{a_1, a_2, a_3, \ldots\}\]A sequence can also be viewed as a function whose domain is the positive integers. For example, if \(a_n = n/(n+1)\), then the sequence begins \(\{1/2,\; 2/3,\; 3/4,\; 4/5, \ldots\}\).
1.1 Limits of Sequences
We say a sequence converges to \(L\) if:
\[\lim_{n \to \infty} a_n = L\]meaning the terms get arbitrarily close to \(L\) as \(n \to \infty\). If no such \(L\) exists, the sequence diverges. A sequence whose terms grow without bound is said to diverge to infinity [Stewart §11.1].
The key bridge between sequence limits and function limits:
Theorem. If \(\lim_{x \to \infty} f(x) = L\) and \(f(n) = a_n\) for each integer \(n\), then \(\lim_{n \to \infty} a_n = L\).
This allows us to apply L’Hôpital’s Rule to sequences — but you must switch from the integer variable \(n\) to the real variable \(x\) before using any derivative rules.
1.2 Bounded Monotonic Sequences
A sequence is monotone if it is always increasing or always decreasing. It is bounded if all terms lie within some fixed interval.
Monotone Convergence Theorem (MCT). Every sequence that is both monotone and bounded must converge.
The MCT is an existence theorem — it tells you a limit exists without telling you what it is. It is the key tool behind the proofs of both the Integral Test and the Alternating Series Test.
2. Infinite Series
Now that we understand sequences, we can tackle infinite series. An infinite series is a sum of all the terms of a sequence:
\[\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots\]The meaning of this sum requires care. We define the \(k\)th partial sum:
\[S_k = \sum_{n=1}^{k} a_n = a_1 + a_2 + \cdots + a_k\]An infinite series is, fundamentally, a sequence of partial sums \(\{S_k\}_{k=1}^{\infty}\). We define its sum as:
\[\sum_{n=1}^{\infty} a_n = \lim_{k \to \infty} S_k\]If this limit is a finite real number, the series converges; otherwise it diverges [Stewart §11.2]. This is in direct analogy with the improper integral:
\[\int_1^{\infty} f(x)\,dx = \lim_{k \to \infty} \int_1^k f(x)\,dx\]2.1 Geometric Series (Must Memorize)
\[\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}, \quad |r| < 1\]| The partial sum simplifies to \(S_k = (a - ar^k)/(1-r)\). Taking \(k \to \infty\) gives the result above when $$ | r | < 1\(. The series diverges for\) | r | \geq 1$$. |
2.2 Telescoping Series
Some series have partial sums that simplify by cancellation. For example:
\[\sum_{n=1}^{\infty} \left(\frac{1}{2^n} - \frac{1}{2^{n+1}}\right) = \lim_{k\to\infty}\left(\frac{1}{2} - \frac{1}{2^{k+1}}\right) = \frac{1}{2}\]2.3 The Divergence Test
Divergence Test. If \(\lim_{n \to \infty} a_n \neq 0\), or if the limit does not exist, then \(\sum a_n\) diverges.
⚠️ The converse is false: \(\lim a_n = 0\) does not guarantee convergence. The harmonic series \(\sum 1/n\) is the classic counterexample — its terms go to zero, yet it diverges.
3. The Integral Test and the \(p\)-Series
3.1 The Integral Test
Setup: Suppose \(f\) is continuous, positive, and decreasing on \([1, \infty)\), and \(a_k = f(k)\) for each \(k\). Then [Stewart §11.3]:
\[\sum_{k=1}^{\infty} a_k \quad \text{and} \quad \int_1^{\infty} f(x)\,dx \quad \text{either both converge or both diverge.}\]The idea is geometric: the terms of the series correspond to rectangle areas that bracket the area under the curve. The Monotone Convergence Theorem is the backbone of the proof.
3.2 The \(p\)-Series (Must Memorize)
\[\sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{converges} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}\]The sum of a \(p\)-series is only known for even integer values \(p = 2, 4, 6, \ldots\) (first discovered by Euler). For example, \(\sum 1/n^2 = \pi^2/6\). The sum \(\sum 1/n^3\) is still unknown.
3.3 Remainder Estimate for the Integral Test
If \(\sum a_n\) converges by the Integral Test and the \(k\)th remainder is \(R_k = a_{k+1} + a_{k+2} + \cdots\), then:
\[\int_{k+1}^{\infty} f(x)\,dx \;\leq\; R_k \;\leq\; \int_k^{\infty} f(x)\,dx\]This also gives an improved two-sided estimate for the total sum:
\[S_k + \int_{k+1}^{\infty} f(x)\,dx \;\leq\; \sum_{n=1}^{\infty} a_n \;\leq\; S_k + \int_k^{\infty} f(x)\,dx\]4. Comparison Tests
4.1 Direct Comparison Test
Suppose \(0 \leq a_n \leq b_n\) for all large \(n\) [Stewart §11.4]:
- If \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(\sum a_n\) diverges, then \(\sum b_n\) diverges.
The intuition comes from the Monotone Convergence Theorem: positive terms give increasing partial sums, and an upper bound prevents them from blowing up.
4.2 Limit Comparison Test
Suppose \(a_n, b_n > 0\) and the following limit exists:
\[\lim_{n \to \infty} \frac{a_n}{b_n} = c, \quad 0 < c < \infty\]Then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge.
Strategy: Identify the dominant terms of \(a_n\) as \(n \to \infty\). If \(a_n \sim 1/n^p\), use Limit Comparison with the \(p\)-series \(\sum 1/n^p\).
⚠️ Important: If the series behaves like a \(p\)-series, the Ratio Test and Root Test will always be inconclusive (giving \(L = 1\)). Always use the Limit Comparison Test in this case.
4.3 Example: When Direct Comparison Is Awkward
Determine whether \(\displaystyle\sum_{n=1}^{\infty} \frac{n+5}{4n^8 - 1}\) converges or diverges.
Direct Comparison is tricky here. Instead, note that for large \(n\), the terms behave like \(1/n^7\). Applying the Limit Comparison Test with \(\sum 1/n^7\):
\[\lim_{n \to \infty} \frac{(n+5)/(4n^8-1)}{1/n^7} = \lim_{n \to \infty} \frac{n^8 + 5n^7}{4n^8 - 1} = \frac{1}{4} > 0\]Since \(\sum 1/n^7\) converges (\(p\)-series with \(p = 7 > 1\)) and the limit is a finite positive number, the series converges.
5. Alternating Series
5.1 Definition and the Alternating Series Test (AST)
A series \(\sum a_n\) is alternating if it can be written as:
\[\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots, \quad b_n > 0\]
Alternating Series Test (AST). Let \(\sum a_n\) be an alternating series. If $${ a_n }\(is decreasing and\)\lim_{n \to \infty} a_n = 0\(, then\)\sum a_n$$ converges [Stewart §11.5].
| The proof uses the Monotone Convergence Theorem applied separately to the even and odd partial sums; the condition $$\lim | a_n | = 0$$ forces both subsequences to the same limit. |
Classic example: The alternating harmonic series
\[\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\]converges by the AST, and its sum equals \(\ln 2\).
| ⚠️ Always check the Divergence Test first. If $$\lim | a_n | \neq 0$$, the series diverges by the Divergence Test — the AST cannot be applied. |
5.2 Alternating Series Estimation Theorem (ASE)
ASE Theorem. If \(\sum a_n\) converges by the AST and the \(k\)th remainder is \(R_k = S - S_k\), then:
\[|R_k| \leq |a_{k+1}|\]The error is bounded by the absolute value of the first omitted term. Moreover, the true sum \(S\) always lies between two consecutive partial sums \(S_k\) and \(S_{k+1}\).
Example: For the alternating harmonic series, to approximate the sum within \(10^{-4}\), we need \(1/(n+1) < 10^{-4}\), giving \(n > 9999\). So \(S_{10000}\) suffices.
5.3 Absolute and Conditional Convergence
\(\sum a_n\) is absolutely convergent if $$\sum a_n $$ converges. \(\sum a_n\) is conditionally convergent if \(\sum a_n\) converges but $$\sum a_n $$ diverges.
Theorem: Absolute convergence implies convergence (but not the other way around).
| Series | \(\sum|a_n|\) | Verdict |
|---|---|---|
| \(\sum (-1)^{n-1}/n^2\) | Converges (\(p\)-series, \(p=2\)) | Absolutely convergent |
| \(\sum (-1)^{n-1}/n\) | Diverges (harmonic) | Conditionally convergent |
6. The Ratio Test and Root Test
6.1 The Ratio Test
The Ratio Test detects whether a series behaves like a geometric series [Stewart §11.6]. Compute:
\[L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|\]| \(L\) | Conclusion |
|---|---|
| \(L < 1\) | Absolutely convergent |
| \(L > 1\) or \(L = \infty\) | Divergent |
| \(L = 1\) | Inconclusive |
Best used when: \(a_n\) involves factorials or exponentials. Never use the Ratio Test when \(a_n\) is a rational or algebraic function of \(n\) — it will always give \(L = 1\).
Example: For \(\displaystyle\sum_{n=1}^{\infty} \frac{(-3)^n}{n!}\), we compute:
\[\left|\frac{a_{n+1}}{a_n}\right| = \frac{3}{n+1} \;\to\; 0 < 1\]So the series converges absolutely.
6.2 The Root Test
\[L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\]Same conclusion table as the Ratio Test. Best used when \(a_n = (f(n))^n\).
7. Strategy for Testing Series
There is no single algorithm, but the following decision process covers most cases [Stewart §11.7]:
- Divergence Test first. If \(\lim a_n \neq 0\), the series diverges. Done.
- Recognize special forms. Geometric? \(p\)-series? Telescoping? Apply the known result directly.
- Rational or algebraic terms? Use Limit Comparison with an appropriate \(p\)-series.
- Factorials or exponentials? Use the Ratio Test.
- Terms of the form \((f(n))^n\)? Use the Root Test.
- Alternating signs? Try the Alternating Series Test.
- Can you bound the terms? Try Direct Comparison.
- Last resort: Integral Test, if \(a_n = f(n)\) with \(f\) continuous, positive, and decreasing.
8. Power Series
8.1 Definition
A power series centered at \(a\) is an infinite series of the form [Stewart §11.8]:
\[\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots\]For each fixed value of \(x\), this is an ordinary infinite series that may converge or diverge. By convention, \(x^0 = 1\) for all \(x\), including \(x = 0\).
8.2 Radius and Interval of Convergence
For any power series, exactly one of three things is true:
- The series converges only at \(x = a\).
- The series converges for all \(x\).
There is a number \(R > 0\) such that the series converges for $$ x - a < R\(and diverges for\) x - a > R$$.
\(R\) is the radius of convergence. The interval of convergence is centered at \(a\) with half-width \(R\). Always check the endpoints separately — the series may converge or diverge there.
To find \(R\): Apply the Ratio Test to the power series treating \(x\) as a fixed but unspecified constant. Solve \(L < 1\) for \(x\).
8.3 Differentiation and Integration of Power Series
If \(f(x) = \sum_{n=0}^{\infty} c_n(x-a)^n\) has radius of convergence \(R > 0\), then \(f\) is differentiable and integrable on \((a-R,\; a+R)\):
\[f'(x) = \sum_{n=1}^{\infty} n\, c_n(x-a)^{n-1}\] \[\int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}\]Both the derivative and antiderivative have the same radius of convergence \(R\) (though endpoints may change).
9. Representations of Functions as Power Series
9.1 Starting From the Geometric Series
The most basic power series is [Stewart §11.9]:
\[\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots, \quad R = 1\]By substitution, differentiation, and integration, we derive power series for many other functions from this single starting point.
9.2 Deriving \(\ln(1+x)\)
Since \(\frac{d}{dx}\ln(1+x) = \frac{1}{1+x}\), and we know:
\[\frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^n x^n, \quad |x| < 1\]integrating term-by-term and using \(\ln(1) = 0\) to find the constant:
\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n, \quad R = 1\]9.3 Deriving \(\arctan x\)
Since \(\frac{d}{dx}\arctan x = \frac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n x^{2n}\), integrating and using \(\arctan(0) = 0\):
\[\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, \quad R = 1\]Power series keep their radius of convergence under integration — this is what makes it possible to evaluate integrals like \(\int \ln(1+x)/x\; dx\) as a power series.
10. Taylor and Maclaurin Series
10.1 Meaning of the Coefficients
Suppose \(f\) is represented by a power series centered at \(a\) with radius \(R > 0\):
\[f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots\]Plugging in \(x = a\) gives \(c_0 = f(a)\). Differentiating and plugging in \(x = a\) gives \(c_1 = f'(a)\). Differentiating twice gives \(2c_2 = f''(a)\). Repeating this process reveals the pattern:
\[c_n = \frac{f^{(n)}(a)}{n!}\]10.2 The Taylor Series
Theorem [Stewart §11.10]. IF \(f\) has a power series representation at \(a\) with radius \(R > 0\), THEN it must be:
\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]
This is the Taylor series of \(f\) centered at \(a\). When \(a = 0\), it is called the Maclaurin series:
\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\]Uniqueness. The representation is unique: for any center \(a\), a function \(f\) has at most one power series representation, which must be its Taylor series.
Two important warnings:
- Warning 1: For some functions, the Taylor series may have radius of convergence zero.
- Warning 2: Even if the radius of convergence is nonzero, the Taylor series may not equal \(f\) everywhere in that interval.
10.3 The Essential Maclaurin Series (Memorize These)
| Function | Maclaurin Series | Radius |
|---|---|---|
| \(e^x\) | \(\displaystyle 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty}\frac{x^n}{n!}\) | \(R = \infty\) |
| \(\sin x\) | \(\displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\) | \(R = \infty\) |
| \(\cos x\) | \(\displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\) | \(R = \infty\) |
| \(1/(1-x)\) | \(\displaystyle 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty} x^n\) | \(R = 1\) |
| \(\ln(1+x)\) | \(\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n\) | \(R = 1\) |
| \(\arctan x\) | \(\displaystyle x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}\) | \(R = 1\) |
| \((1+x)^k\) | \(\displaystyle 1 + kx + \frac{k(k-1)}{2!}x^2 + \cdots = \sum_{n=0}^{\infty}\binom{k}{n}x^n\) | \(R = 1\) |
10.4 Taylor Polynomials
Each partial sum of the Taylor series is a polynomial. The \(k\)th degree Taylor polynomial of \(f\) centered at \(a\) is:
\[T_k(x) = \sum_{n=0}^{k} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k\]\(T_k\) is the best degree-\(k\) polynomial approximation to \(f\) near \(x = a\). As \(k\) increases, \(T_k\) hugs the curve more closely over a wider interval.
10.5 Taylor’s Remainder Formula (Taylor’s Inequality)
| The error in using \(T_k(x_0)\) to approximate \(f(x_0)\) is \(R_k(x_0) = f(x_0) - T_k(x_0)\). If $$ | f^{(k+1)}(z) | \leq M\(for all\)z\(between\)x_0\(and\)a$$, then: |
| For \(f(x) = \sin x\) or \(\cos x\), all derivatives satisfy $$ | f^{(k+1)}(z) | \leq 1\(, so\)M = 1\(always works. Using this inequality, one can prove that\)e^x\(,\)\sin x\(, and\)\cos x\(are each equal to their Maclaurin series for all\)x$$. |
If $$ R_k(x_0) \to 0\(as\)k \to \infty\(, then the Taylor series truly equals\)f(x_0)$$ at that point.
11. Applications of Taylor Series
With the Maclaurin series table in hand, Taylor series become a powerful tool for problems that ordinary calculus cannot handle directly [Stewart §11.11].
11.1 Evaluating Integrals Without Closed Forms
Some integrands have no elementary antiderivative. Substitute a known Maclaurin series and integrate term-by-term:
\[\int e^{-x^2}\,dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}\,dx = C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)}\]For a definite integral, substitute the upper limit and use the ASE Theorem to bound the error from truncating the series.
11.2 Identifying the Sum of a Series
Recognize a given series as a known Maclaurin series. For example:
\[7 - \frac{7}{2!} + \frac{7}{3!} - \frac{7}{4!} + \cdots = 7\sum_{n=0}^{\infty}\frac{(-1)^n}{n!} = 7e^{-1} = \frac{7}{e}\]11.3 Evaluating Limits Using Taylor Series
For indeterminate forms, substituting Taylor series is often far cleaner than repeated L’Hôpital’s Rule. For example:
\[\lim_{x \to 0} \frac{\sin x - x}{x^3}\]Substituting the Maclaurin series for \(\sin x\):
\[= \lim_{x \to 0} \frac{\left(x - \dfrac{x^3}{6} + \cdots\right) - x}{x^3} = \lim_{x \to 0} \frac{-\dfrac{x^3}{6} + \cdots}{x^3} = -\frac{1}{6}\]11.4 Approximating Functions and Bounding Errors
Find the 2nd degree Taylor polynomial for \(f(x) = x^{2/3}\) centered at \(a = 8\):
\[T_2(x) = 4 + \frac{1}{3}(x-8) - \frac{1}{72}(x-8)^2\]Use it to estimate \(f(8.1) \approx T_2(8.1)\), then apply Taylor’s Inequality to bound the error.
12. Summary
| Concept | Key Formula or Fact | ||
|---|---|---|---|
| Sequence limit | \(\lim_{n\to\infty} a_n = L\); apply L’Hôpital via \(f(x)\) if needed | ||
| Monotone Convergence Thm | Bounded + monotone \(\Rightarrow\) convergent | ||
| Geometric series | \(\sum ar^{n-1} = a/(1-r)\) for \(|r| < 1\) | ||
| Divergence Test | \(\lim a_n \neq 0 \Rightarrow\) diverges | ||
| \(p\)-series | Converges iff \(p > 1\) | ||
| Integral Test | \(\sum a_n\) and \(\int f\,dx\) share convergence/divergence | ||
| Limit Comparison Test | \(\lim a_n/b_n = c \in (0,\infty) \Rightarrow\) same behavior | ||
| AST | $$ | a_n | \(decreasing to 0\)\Rightarrow$$ alternating series converges |
| ASE Theorem | \(|R_k| \leq |a_{k+1}|\) | ||
| Ratio Test | \(L = \lim |a_{n+1}/a_n|\); converges abs. if \(L < 1\) | ||
| Power series | Radius of convergence via Ratio or Root Test | ||
| Taylor series | \(\sum f^{(n)}(a)(x-a)^n/n!\); unique if it exists | ||
| Taylor’s Inequality | \(|R_k(x_0)| \leq \frac{M}{(k+1)!}|x_0-a|^{k+1}\) |
The deepest payoff of Chapter 11 is the realization that familiar functions like \(e^x\), \(\sin x\), and \(\cos x\) are infinite polynomials — and that fact alone lets us integrate them over any interval, approximate them to any precision, and compute their values from scratch using only addition and multiplication.
References
All section numbers refer to:
Stewart, J. (2021). Calculus, 9th edition. Cengage Learning.
- §11.1 — Sequences
- §11.2 — Series
- §11.3 — The Integral Test and Estimates of Sums
- §11.4 — The Comparison Tests
- §11.5 — Alternating Series and Absolute Convergence
- §11.6 — The Ratio and Root Tests
- §11.7 — Strategy for Testing Series
- §11.8 — Power Series
- §11.9 — Representations of Functions as Power Series
- §11.10 — Taylor and Maclaurin Series
- §11.11 — Applications of Taylor Polynomials
