Infinite Series Convergence & Power Series
ยง7.1 Infinite Sequences, Series & The Cauchy Convergence Criterion
1. Sequences and the Sequence of Partial Sums
An infinite sequence is an ordered map $a: \mathbb{N} \to \mathbb{R}$, denoted $(a_n)_{n=1}^\infty$. An infinite series is the formal sum $\sum_{n=1}^\infty a_n$.
Its convergence is defined strictly as the limit of its sequence of partial sums $s_N = \sum_{n=1}^N a_n$:
If $\lim_{N \to \infty} s_N$ does not exist as a finite real number, the series diverges.
2. The Necessary Divergence Test ($n$-th Term Test)
Contrapositive: If $\lim_{n \to \infty} a_n \ne 0$ (or does not exist), the series $\sum a_n$ diverges. Warning: The converse is false; the Harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ satisfies $\lim \frac{1}{n} = 0$ but diverges logarithmically.
3. The Cauchy Criterion for Series
This characterization relies solely on the completeness of $\mathbb{R}$, eliminating the need to know the sum $S$ a priori.
ยง7.2 Convergence Tests for Non-Negative Series: Integral, Comparison, Ratio & Root
1. The Integral Test (Maclaurin-Cauchy)
Let $f: [1, \infty) \to \mathbb{R}$ be continuous, positive, and monotonically decreasing such that $f(n) = a_n$. Then:
Application to $p$-Series: $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$.
2. Direct Comparison & Limit Comparison Tests
- Direct: If $0 \le a_n \le b_n$, then $\sum b_n < \infty \implies \sum a_n < \infty$; and $\sum a_n = \infty \implies \sum b_n = \infty$.
- Limit: If $a_n > 0, b_n > 0$ and $L = \lim_{n \to \infty} \frac{a_n}{b_n} \in (0, \infty)$, then $\sum a_n$ and $\sum b_n$ share the exact same convergence status.
3. d'Alembert's Ratio Test
Let $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$:
- If $L < 1$: The series converges absolutely.
- If $L > 1$: The series diverges.
- If $L = 1$: The test is inconclusive (requires Raabe's test or Integral test).
4. Cauchy's Root Test
Let $L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$:
- If $L < 1$: Converges absolutely.
- If $L > 1$: Diverges.
- If $L = 1$: Inconclusive.
ยง7.3 Alternating Series, Absolute vs Conditional Convergence & Riemann Rearrangement
1. The Leibniz Alternating Series Test
An alternating series $\sum_{n=1}^\infty (-1)^{n-1} b_n$ with $b_n > 0$ converges if:
- $b_{n+1} \le b_n$ for all $n \ge N$ (monotonic decrease).
- $\lim_{n \to \infty} b_n = 0$.
Alternating Series Estimation Theorem: If $S$ is the sum, the truncation error $|R_N| = |S - s_N|$ satisfies:
The error is strictly bounded by the magnitude of the first neglected term.
2. Absolute vs Conditional Convergence
- Absolute Convergence: $\sum |a_n|$ converges. Every absolutely convergent series converges.
- Conditional Convergence: $\sum a_n$ converges, but $\sum |a_n| = \infty$ (e.g. the alternating harmonic series $\sum \frac{(-1)^{n-1}}{n} = \ln 2$).
3. Riemann's Rearrangement Theorem
If a series $\sum a_n$ is conditionally convergent, then for any real number $M \in \mathbb{R}$ (or $\pm\infty$), there exists a permutation $\sigma: \mathbb{N} \to \mathbb{N}$ of the series indices such that the rearranged series sums exactly to $M$:
In contrast, absolutely convergent series can be rearranged arbitrarily without altering their sum (Dirichlet's theorem).
ยง7.4 Power Series & The Cauchy-Hadamard Radius of Convergence
1. Definition of a Power Series
A power series centered at $x_0 \in \mathbb{R}$ with real coefficients $(c_n)$ is:
2. The Cauchy-Hadamard Theorem
- The series converges absolutely for all $|x - x_0| < R$.
- The series diverges for all $|x - x_0| > R$.
- At the boundary points $x = x_0 \pm R$, the series may converge absolutely, converge conditionally, or diverge.
The Radius of Convergence $R$ is determined by the Cauchy-Hadamard formula:
The Interval of Convergence is one of $(x_0 - R, x_0 + R)$, $[x_0 - R, x_0 + R)$, $(x_0 - R, x_0 + R]$, or $[x_0 - R, x_0 + R]$.
ยง7.5 Term-by-Term Differentiation & Integration of Power Series
1. Uniform Convergence on Compact Subsets
Within its open disk of convergence $|x - x_0| < R$, a power series converges uniformly on every compact subinterval $[x_0 - r, x_0 + r]$ ($r < R$). Consequently, $f(x)$ is infinitely differentiable ($C^\infty$) on $(x_0 - R, x_0 + R)$.
2. Term-by-Term Differentiation Theorem
The derivative of $f(x) = \sum_{n=0}^\infty c_n (x - x_0)^n$ is obtained by differentiating term by term:
The derivative series has the exact same radius of convergence $R$.
3. Term-by-Term Integration Theorem
The indefinite integral of $f(x)$ is obtained by integrating term by term:
The integrated series also preserves the exact same radius of convergence $R$.
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
Find the radius and exact interval of convergence of the power series, testing both boundary endpoints:
Step 1: Compute the Radius of Convergence $R$ via the Ratio Test
Let $u_n(x) = \frac{(-1)^n (x - 3)^n}{n \cdot 5^n}$. Compute the consecutive ratio:
Taking the limit as $n \to \infty$:
For absolute convergence, $L < 1 \implies \frac{|x - 3|}{5} < 1 \implies |x - 3| < 5$.
Thus, the Radius of Convergence is $R = 5$, and the open interval is $(-2, 8)$.
Step 2: Test the Left Endpoint $x = -2$
Substitute $x = -2 \implies x - 3 = -5$:
This is the standard Harmonic Series, which diverges ($p = 1$). Hence $x = -2$ is excluded.
Step 3: Test the Right Endpoint $x = 8$
Substitute $x = 8 \implies x - 3 = 5$:
This is the Alternating Harmonic Series, which converges by the Leibniz test. Hence $x = 8$ is included.
Conclusion
$R = 5$; Interval of Convergence is $(-2, 8]$.
Evaluate the exact sum of the series using term-by-term differentiation of the geometric series:
Step 1: Start with the Geometric Series
For $|x| < 1$, the geometric series evaluates to:
Step 2: Differentiate Term-by-Term
Differentiating both sides with respect to $x$:
Step 3: Multiply by $x$
Step 4: Substitute $x = 1/2$
Since $|1/2| < 1$, the substitution is valid:
$S = 2$
Let $\sum_{n=0}^\infty c_n x^n$ be a power series. Define $\rho = \limsup_{n \to \infty} \sqrt[n]{|c_n|}$ and $R = 1/\rho$ (with $R = \infty$ if $\rho = 0$, and $R = 0$ if $\rho = \infty$). Prove that: (a) The series converges absolutely for all $|x| < R$. (b) The series diverges for all $|x| > R$.
Part (a): Proof of Absolute Convergence for $|x| < R$
Assume $0 < \rho < \infty$ and let $|x| < R = 1/\rho$, so $|x|\rho < 1$. Choose $\varepsilon > 0$ small enough such that:
By the definition of the limit superior $\limsup \sqrt[n]{|c_n|} = \rho$, there exists an integer $N \in \mathbb{N}$ such that for all $n \ge N$:
Multiplying by $|x|^n$:
Since $q < 1$, the geometric series $\sum_{n=N}^\infty q^n$ converges. By the Direct Comparison Test, $\sum_{n=0}^\infty |c_n x^n|$ converges, establishing that $\sum c_n x^n$ converges absolutely for all $|x| < R$. $\blacksquare$
Part (b): Proof of Divergence for $|x| > R$
Assume $|x| > R = 1/\rho$, so $|x|\rho > 1$. Then $\rho > 1/|x|$.
By the definition of the limit superior, there exist infinitely many indices $n_k \in \mathbb{N}$ such that:
Thus, the sequence of terms $a_n = c_n x^n$ does NOT converge to zero as $n \to \infty$ ($\lim_{n \to \infty} c_n x^n \ne 0$).
By the $n$-th term divergence test, the series $\sum c_n x^n$ must diverge. $\blacksquare$
Proved: $\sum c_n x^n$ converges absolutely for $|x| < R$ and diverges for $|x| > R$ where $1/R = \limsup \sqrt[n]{|c_n|}$.