Mathematics / Analysis Integral Calculus & Series 100% Free Open Access
Chapter 7 โ€ข Theory & Derivations

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 $s_N = \sum_{n=1}^N a_n$:

$$\sum_{n=1}^\infty a_n = \lim_{N \to \infty} s_N = S \in \mathbb{R}$$

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)

$$\mathbf{\sum_{n=1}^\infty a_n \text{ converges} \implies \lim_{n \to \infty} a_n = 0}$$

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

$$\mathbf{\sum_{n=1}^\infty a_n \text{ converges} \iff \forall \varepsilon > 0, \exists N \in \mathbb{N} \text{ such that } \forall m > n \ge N: \left| \sum_{k=n+1}^m a_k \right| < \varepsilon}$$

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:

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

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:

  1. $b_{n+1} \le b_n$ for all $n \ge N$ (monotonic decrease).
  2. $\lim_{n \to \infty} b_n = 0$.

Alternating Series Estimation Theorem: If $S$ is the sum, the truncation error $|R_N| = |S - s_N|$ satisfies:

$$|S - s_N| \le b_{N+1}$$

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 $M$:

$$\sum_{n=1}^\infty a_{\sigma(n)} = M$$

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:

$$f(x) = \sum_{n=0}^\infty c_n (x - x_0)^n = c_0 + c_1(x - x_0) + c_2(x - x_0)^2 + \dots$$

2. The Cauchy-Hadamard Theorem

$$\mathbf{\text{Theorem: For any power series } \sum c_n (x - x_0)^n, \text{ there exists a unique } R \in [0, \infty] \text{ such that:}}$$
  • 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:

$$\mathbf{\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|c_n|} \quad \text{or} \quad R = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right|}$$

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:

$$\mathbf{f'(x) = \sum_{n=1}^\infty n c_n (x - x_0)^{n-1} = \sum_{k=0}^\infty (k+1) c_{k+1} (x - x_0)^k}$$

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:

$$\mathbf{\int f(x) \, dx = C + \sum_{n=0}^\infty \frac{c_n}{n+1} (x - x_0)^{n+1}}$$

The integrated series also preserves the exact same radius of convergence $R$.

TIERED UNIVERSITY HONORS PROBLEMS

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.

Tier 1: Foundational Example 7.1: Interval of Convergence with Boundary Endpoint Testing

Find the radius and exact interval of convergence of the power series, testing both boundary endpoints:

$$\sum_{n=1}^\infty \frac{(-1)^n (x - 3)^n}{n \cdot 5^n}$$

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:

$$\left| \frac{u_{n+1}(x)}{u_n(x)} \right| = \left| \frac{(x-3)^{n+1}}{(n+1) 5^{n+1}} \cdot \frac{n \cdot 5^n}{(x-3)^n} \right| = \frac{|x - 3|}{5} \cdot \frac{n}{n+1}$$

Taking the limit as $n \to \infty$:

$$L = \lim_{n \to \infty} \left| \frac{u_{n+1}}{u_n} \right| = \frac{|x - 3|}{5} \lim_{n \to \infty} \frac{n}{n+1} = \frac{|x - 3|}{5}$$

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 $x = -2 \implies x - 3 = -5$:

$$\sum_{n=1}^\infty \frac{(-1)^n (-5)^n}{n \cdot 5^n} = \sum_{n=1}^\infty \frac{(-1)^n (-1)^n 5^n}{n \cdot 5^n} = \sum_{n=1}^\infty \frac{1}{n}$$

This is the standard Harmonic Series, which diverges ($p = 1$). Hence $x = -2$ is excluded.

Step 3: Test the Right Endpoint $x = 8$

Substitute $x = 8 \implies x - 3 = 5$:

$$\sum_{n=1}^\infty \frac{(-1)^n 5^n}{n \cdot 5^n} = \sum_{n=1}^\infty \frac{(-1)^n}{n}$$

This is the Alternating Harmonic Series, which converges by the Leibniz test. Hence $x = 8$ is included.

Conclusion

$$\mathbf{R = 5, \qquad \text{Interval of Convergence: } (-2, 8]}$$
Final Answer & Physical Insight

$R = 5$; Interval of Convergence is $(-2, 8]$.

Tier 2: Intermediate Exam Example 7.2: Summation of Series via Term-by-Term Differentiation

Evaluate the exact sum of the series using term-by-term differentiation of the geometric series:

$$S = \sum_{n=1}^\infty \frac{n}{2^n}$$

Step 1: Start with the Geometric Series

For $|x| < 1$, the geometric series evaluates to:

$$\sum_{n=0}^\infty x^n = \frac{1}{1 - x}$$

Step 2: Differentiate Term-by-Term

Differentiating both sides with respect to $x$:

$$\frac{d}{dx} \left( \sum_{n=0}^\infty x^n \right) = \sum_{n=1}^\infty n x^{n-1} = \frac{d}{dx} (1 - x)^{-1} = \frac{1}{(1 - x)^2}$$

Step 3: Multiply by $x$

$$\sum_{n=1}^\infty n x^n = x \sum_{n=1}^\infty n x^{n-1} = \frac{x}{(1 - x)^2}$$

Step 4: Substitute $x = 1/2$

Since $|1/2| < 1$, the substitution is valid:

$$S = \sum_{n=1}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty n \left(\frac{1}{2}\right)^n = \frac{1/2}{\left(1 - \frac{1}{2}\right)^2} = \frac{1/2}{(1/2)^2} = \frac{1/2}{1/4} = \mathbf{2}$$
Final Answer & Physical Insight

$S = 2$

Tier 3: Honors / Proof Challenge Example 7.3: Rigorous Proof of the Cauchy-Hadamard Theorem

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:

$$q = |x|(\rho + \varepsilon) < 1$$

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 $n \ge N$:

$$\sqrt[n]{|c_n|} < \rho + \varepsilon \implies |c_n| < (\rho + \varepsilon)^n$$

Multiplying by $|x|^n$:

$$|c_n x^n| = |c_n| |x|^n < \Big( |x|(\rho + \varepsilon) \Big)^n = q^n$$

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:

$$\sqrt[n_k]{|c_{n_k}|} > \frac{1}{|x|} \implies |c_{n_k}| > \left(\frac{1}{|x|}\right)^{n_k} \implies |c_{n_k} x^{n_k}| = |c_{n_k}| |x|^{n_k} > 1$$

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. $\blacksquare$

Final Answer & Physical Insight

Proved: $\sum c_n x^n$ converges absolutely for $|x| < R$ and diverges for $|x| > R$ where $1/R = \limsup \sqrt[n]{|c_n|}$.