Unit 4: Infinite Series of Real Numbers: Rigorous Convergence Tests

1. The Concept of an Infinite Series

An infinite series is the formal addition of countably infinitely many real numbers:

Because infinity is not a real number, addition cannot be performed term-by-term infinitely. Instead, analysis rigorously reduces the convergence of an infinite series to the convergence of a sequence of finite partial sums.

Definition 4.1 (Partial Sums and Series Convergence): Given a sequence of real numbers $(a_n)_{n=1}^\infty$, the $k$-th partial sum is:

The infinite series $\sum_{n=1}^\infty a_n$ is said to converge to the sum $S \in \mathbb{R}$ if the sequence of partial sums $(s_k)_{k=1}^\infty$ converges to $S$:

If $(s_k)$ diverges, the series is said to diverge.

2. The $n$-th Term Test for Divergence

Theorem 4.1 ($n$-th Term Divergence Test): If the series $\sum_{n=1}^\infty a_n$ converges, then:

Equivalently, if $\lim_{n \to \infty} a_n \ne 0$ or does not exist, the series diverges.

Proof:

Suppose $\sum_{n=1}^\infty a_n = S$. Then $\lim_{k \to \infty} s_k = S$ and $\lim_{k \to \infty} s_{k-1} = S$. For any $n \ge 2$, the $n$-th term is the difference between consecutive partial sums:

By the algebraic limit theorem for differences:

Warning (Non-Sufficiency of $a_n \to 0$): The condition $\lim a_n = 0$ is strictly necessary, but not sufficient for convergence! The canonical counterexample is the Harmonic Series:

Here $a_n = \frac{1}{n} \to 0$, yet the series diverges to $\infty$.

3. The Cauchy Criterion for Series

Theorem 4.2 (Cauchy Convergence Criterion for Series): A series $\sum_{n=1}^\infty a_n$ converges in $\mathbb{R}$ if and only if for every $\epsilon > 0$, there exists an index $N \in \mathbb{N}$ such that:

Proof:

The series converges if and only if the sequence of partial sums $(s_k)$ converges. By the Cauchy Completeness Theorem (Theorem 3.10), $(s_k)$ converges in $\mathbb{R}$ if and only if $(s_k)$ is a Cauchy sequence:

Since $s_m - s_n = \sum_{k=n+1}^m a_k$, the assertion is proven. $\blacksquare$

1. Monotonicity of Partial Sums

For series where all terms are non-negative ($a_n \ge 0$), the sequence of partial sums is monotonically increasing:

By the Monotone Convergence Theorem, a non-negative series converges if and only if its sequence of partial sums is bounded above.

2. Direct and Limit Comparison Tests

Theorem 4.3 (Direct Comparison Test): Let $0 \le a_n \le b_n$ for all $n \ge N_0$.

1. If $\sum_{n=1}^\infty b_n$ converges, then $\sum_{n=1}^\infty a_n$ converges.
2. If $\sum_{n=1}^\infty a_n$ diverges, then $\sum_{n=1}^\infty b_n$ diverges.

Theorem 4.4 (Limit Comparison Test): Let $a_n > 0$ and $b_n > 0$ for all $n$, and suppose:

1. If $0 < L < \infty$, then either both series converge or both diverge.
2. If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
3. If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

Rigorous Proof of Case 1 ($0 < L < \infty$):

Let $\epsilon = \frac{L}{2} > 0$. By definition of limit, there exists $N \in \mathbb{N}$ such that for all $n \ge N$:

Multiplying through by $b_n > 0$:

• If $\sum b_n$ converges, then $\sum \frac{3L}{2} b_n$ converges. By the Direct Comparison Test, $\sum a_n$ converges.
• If $\sum b_n$ diverges, then $\sum \frac{L}{2} b_n$ diverges. By the Direct Comparison Test, $\sum a_n$ diverges. $\blacksquare$

3. The Cauchy Condensation Test

Theorem 4.5 (Cauchy Condensation Test): Let $(a_n)_{n=1}^\infty$ be a monotonically decreasing sequence of non-negative real numbers ($a_1 \ge a_2 \ge a_3 \ge \dots \ge 0$). Then the series $\sum_{n=1}^\infty a_n$ converges if and only if the "condensed" series:

converges.

Line-by-Line Proof:

Let $s_n = \sum_{j=1}^n a_j$ and $t_k = \sum_{j=0}^k 2^j a_{2^j}$. Both $(s_n)$ and $(t_k)$ are monotonically increasing sequences.

Part 1: If $\sum 2^k a_{2^k}$ converges, then $\sum a_n$ converges. For any $n \le 2^k - 1$, group the terms of $s_n$ into dyadic blocks:

Thus the sequence of partial sums $(s_n)$ is bounded above by $M$. By MCT, $\sum a_n$ converges.

Part 2: If $\sum a_n$ converges, then $\sum 2^k a_{2^k}$ converges. For any $k$, consider the partial sum $s_{2^k}$:

Thus $t_k \le 2 s_{2^k} \le 2 \sum_{n=1}^\infty a_n < \infty$. Hence $(t_k)$ is bounded above, so the condensed series converges. $\blacksquare$

Application: The $p$-Series Theorem: The series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$. Proof: By condensation, the condensed series is $\sum 2^k \frac{1}{(2^k)^p} = \sum (2^{1-p})^k$, which is a geometric series with ratio $r = 2^{1-p}$. This converges iff $r < 1 \iff 1 - p < 0 \iff p > 1$. $\blacksquare$

1. Absolute vs. Conditional Convergence

Definition 4.2 (Absolute Convergence): A series $\sum_{n=1}^\infty a_n$ is said to converge absolutely if the series of absolute values converges:

If $\sum a_n$ converges but $\sum |a_n|$ diverges, the series is said to converge conditionally.

Theorem 4.6 (Absolute Convergence Implies Convergence): If $\sum_{n=1}^\infty |a_n|$ converges, then $\sum_{n=1}^\infty a_n$ converges.

Proof: By the Cauchy Criterion for series, since $\sum |a_n|$ converges:

By the generalized triangle inequality:

By Theorem 4.2, $\sum a_n$ converges. $\blacksquare$

2. The Cauchy Root Test and d'Alembert Ratio Test

Theorem 4.7 (Cauchy Root Test): Let $\alpha = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$.

1. If $\alpha < 1$, the series $\sum a_n$ converges absolutely.
2. If $\alpha > 1$, the series $\sum a_n$ diverges.
3. If $\alpha = 1$, the test is inconclusive.

Rigorous Proof of Case 1 ($\alpha < 1$):

Since $\alpha < 1$, choose $r$ such that $\alpha < r < 1$. Let $\epsilon = r - \alpha > 0$. By the definition of $\limsup$, there exists $N \in \mathbb{N}$ such that for all $n \ge N$:

Since $0 < r < 1$, the geometric series $\sum_{n=N}^\infty r^n$ converges. By the Direct Comparison Test, $\sum |a_n|$ converges, so $\sum a_n$ converges absolutely. $\blacksquare$

Theorem 4.8 (d'Alembert Ratio Test): Let $a_n \ne 0$ and suppose $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$.

1. If $L < 1$, the series converges absolutely.
2. If $L > 1$, the series diverges.
3. If $L = 1$, the test is inconclusive.

3. The Maclaurin-Cauchy Integral Test

Theorem 4.9 (The Integral Test): Let $f: [1, \infty) \to [0, \infty)$ be a continuous, non-negative, and monotonically decreasing function such that $f(n) = a_n$ for all $n \in \mathbb{N}$. Then the series $\sum_{n=1}^\infty a_n$ converges if and only if the improper Riemann integral converges:

Proof Sketch: On each interval $[k, k+1]$, since $f$ is decreasing:

Summing from $k=1$ to $n-1$:

The partial sums are bounded if and only if the improper integral is bounded. $\blacksquare$

1. When the Ratio Test Fails ($L = 1$)

When $\lim |a_{n+1}/a_n| = 1$, the Ratio Test provides no information. This occurs for all $p$-series, since $\frac{n^p}{(n+1)^p} = (1 + 1/n)^{-p} \to 1$. To handle these delicate boundary cases, higher-order asymptotic tests are required.

2. Raabe's Test

Theorem 4.10 (Raabe's Test): Let $a_n > 0$ and suppose:

1. If $R > 1$, the series $\sum a_n$ converges.
2. If $R < 1$, the series $\sum a_n$ diverges.
3. If $R = 1$, the test is inconclusive.

Proof Sketch:

Compare $a_n$ against the $p$-series $b_n = \frac{1}{n^p}$. Using the binomial expansion:

So $n \left( \frac{b_n}{b_{n+1}} - 1 \right) \to p$. If $R > 1$, choose $p$ such that $R > p > 1$. For large $n$, $\frac{a_n}{a_{n+1}} > \frac{b_n}{b_{n+1}}$, which forces $\sum a_n$ to converge by comparison with the convergent $p$-series. $\blacksquare$

3. Gauss's Test

Theorem 4.11 (Gauss's Hypergeometric Test): Let $a_n > 0$. Suppose the ratio can be expanded asymptotically as:

where $\delta > 0$ and $(B_n)$ is a bounded sequence. Then:

1. The series converges if $h > 1$.
2. The series diverges if $h \le 1$.

Note that Gauss's test completely resolves the critical boundary $h = 1$ where Raabe's test is inconclusive!

1. The Leibniz Alternating Series Test

Theorem 4.12 (Leibniz Alternating Series Test): Let $(b_n)_{n=1}^\infty$ be a sequence such that:

1. $b_n > 0$ for all $n$.
2. $b_{n+1} \le b_n$ for all $n$ (monotonically decreasing).
3. $\lim_{n \to \infty} b_n = 0$.

Then the alternating series $\sum_{n=1}^\infty (-1)^{n+1} b_n = b_1 - b_2 + b_3 - b_4 + \dots$ converges. Furthermore, the remainder satisfies $|R_n| = |S - s_n| \le b_{n+1}$.

2. Dirichlet's and Abel's Tests for Convergence

Both tests rely on the technique of Abel summation by parts:

Theorem 4.13 (Dirichlet's Test): If the partial sums $s_n = \sum_{k=1}^n a_k$ are bounded (i.e., $|s_n| \le M$ for all $n$), and $(b_n)$ decreases monotonically to 0 ($b_n \ge b_{n+1} \ge 0$ and $b_n \to 0$), then $\sum_{n=1}^\infty a_n b_n$ converges.

Theorem 4.14 (Abel's Test): If $\sum a_n$ converges, and $(b_n)$ is a monotone bounded sequence, then $\sum a_n b_n$ converges.

3. Riemann's Rearrangement Theorem

One of the most astonishing theorems in classical analysis, demonstrating the fundamental fragility of conditional convergence:

Theorem 4.15 (Riemann Rearrangement Theorem): Let $\sum_{n=1}^\infty a_n$ be a conditionally convergent series of real numbers. For any extended real number $S \in [-\infty, \infty]$, there exists a bijection (permutation) $\sigma: \mathbb{N} \to \mathbb{N}$ such that the rearranged series:

(In stark contrast, if $\sum a_n$ converges absolutely, every rearrangement converges to the exact same sum).

Line-by-Line Proof:

Decompose each term into positive and negative parts:

So $a_n = p_n - q_n$ and $|a_n| = p_n + q_n$. Since $\sum a_n$ converges conditionally:

• $\sum a_n = \sum (p_n - q_n)$ converges.
• $\sum |a_n| = \sum (p_n + q_n)$ diverges.

If either $\sum p_n$ or $\sum q_n$ converged, the other would have to converge as well, forcing $\sum |a_n|$ to converge. Thus, both $\sum p_n = \infty$ and $\sum q_n = \infty$ must diverge to $+\infty$! Furthermore, since $\sum a_n$ converges, $\lim p_n = 0$ and $\lim q_n = 0$.

Let $S \in \mathbb{R}$ be any target number.

1. Take positive terms $p_1, p_2, \dots, p_{k_1}$ in original order just until the partial sum strictly exceeds $S$:

(This is always possible because $\sum p_n = \infty$).

1. Now take negative terms $-q_1, -q_2, \dots, -q_{m_1}$ in original order just until the total sum falls strictly below $S$:

(This is always possible because $\sum q_n = \infty$).

1. Continue alternating: add subsequent positive terms until exceeding $S$, then subsequent negative terms until falling below $S$.

At each turn, the difference between the partial sum and $S$ is bounded by the last term added:

Since $p_n \to 0$ and $q_n \to 0$, the overshoot and undershoot shrink to 0 as $N \to \infty$. Therefore, the rearranged series converges precisely to $S$! $\blacksquare$