Series | Mathematics - Wyatt's Notes

A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $S_N = \sum_{n=1}^{N} a_n$ Converges. The limit is the sum of the series.

If $a_n \geq 0$ for all $n$The series of partial sums is increasing, so by the monotone convergence Theorem, $\sum a_n$ converges if and only if $(S_N)$ is bounded above.

Theorem 3.1 (Comparison Test). If $0 \leq a_n \leq b_n$ for all $n$ Then:

• If $\sum b_n$ converges, then $\sum a_n$ converges.
• If $\sum a_n$ diverges, then $\sum b_n$ diverges.

Theorem 3.2 (Limit Comparison Test). If $a_n > 0$, $b_n > 0$ And $\lim_{n \to \infty} a_n / b_n = L$ where $0 \lt L \lt \infty$ Then $\sum a_n$ converges if and only if $\sum b_n$ converges.

Theorem 3.3 (Ratio Test). If $\lim_{n \to \infty} |a_{n+1}/a_n| = L$ Then:

• If $L \lt 1$, $\sum a_n$ converges absolutely.
• If $L > 1$, $\sum a_n$ diverges.
• If $L = 1$The test is inconclusive.

Theorem 3.4 (Root Test). If $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = L$ Then:

• If $L \lt 1$, $\sum a_n$ converges absolutely.
• If $L > 1$, $\sum a_n$ diverges.
• If $L = 1$The test is inconclusive.

Proof. If $L \lt 1$Choose $r$ with $L \lt r \lt 1$. By definition of $\limsup$There exists $N$ such that $\sqrt[n]{|a_n|} \lt r$ for all $n \geq N$I.e., $|a_n| \lt r^n$. Since $\sum r^n$ converges (geometric series with $r \lt 1$), the comparison test gives absolute convergence.

If $L > 1$ Then for infinitely many $n$: $\sqrt[n]{|a_n|} > 1$ So $|a_n| > 1$. Hence $a_n \not\to 0$ And the series diverges. $\blacksquare$

Theorem 3.5 (Integral Test). If $f : [1, \infty) \to [0, \infty)$ is positive, continuous, and Decreasing, then $\sum_{n=1}^{\infty} f(n)$ converges if and only if $\int_1^{\infty} f(x)\, dx$ converges.

Proof. Since $f$ is decreasing, for $k \leq x \leq k+1$: $f(k+1) \leq f(x) \leq f(k)$. Integrating:

$f(k+1) = \int_k^{k+1} f(k+1)\, dx \leq \int_k^{k+1} f(x)\, dx \leq \int_k^{k+1} f(k)\, dx = f(k)$

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

$\sum_{k=2}^{N} f(k) \leq \int_1^N f(x)\, dx \leq \sum_{k=1}^{N-1} f(k)$

If $\int_1^{\infty} f$ converges, the left inequality shows $\sum f(k)$ is bounded above, hence converges. If $\int_1^{\infty} f$ diverges, the right inequality shows $\sum f(k)$ is unbounded, hence diverges. $\blacksquare$

Theorem 3.6 (Alternating Series Test). If $a_n > 0$, $a_n$ decreases, and $a_n \to 0$ Then $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ converges.

Proof. The partial sums of the even-indexed subsequence satisfy $S_{2n} = S_{2n-2} - a_{2n-1} + a_{2n}$. Since $a_{2n-1} \geq a_{2n}$We have $S_{2n} \leq S_{2n-2}$ So $(S_{2n})$ is decreasing. Similarly, $(S_{2n+1})$ is increasing. Also $S_{2n+1} = S_{2n} + a_{2n+1} \geq S_{2n}$. Both sequences are bounded (since $(S_{2n})$ is decreasing and bounded below by $S_1$ And $(S_{2n+1})$ is increasing and bounded Above by $S_2$). Hence both converge. Since $a_{2n+1} \to 0$Their limits coincide. $\blacksquare$

A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. It converges conditionally If $\sum a_n$ converges but $\sum |a_n|$ diverges.

Theorem 3.7. If $\sum a_n$ converges absolutely, then $\sum a_n$ converges.

Proof. Since $\sum |a_n|$ converges, the partial sums of $\sum |a_n|$ satisfy the Cauchy criterion. Given $\varepsilon > 0$There exists $N$ such that for $m > n \geq N$: $\sum_{k=n+1}^{m} |a_k| \lt \varepsilon$. Then $\left|\sum_{k=n+1}^{m} a_k\right| \leq \sum_{k=n+1}^{m} |a_k| \lt \varepsilon$ So $\sum a_n$ satisfies The Cauchy criterion and converges. $\blacksquare$

Theorem 3.8 (Alternating Series Estimation). If $S = \sum_{n=1}^{\infty} (-1)^{n+1} a_n$ satisfies the Hypotheses of the alternating series test, then the error after $N$ terms satisfies:

$|S - S_N| \leq a_{N+1}$

Proof. We have $S_{2n} \leq S \leq S_{2n+1} = S_{2n} + a_{2n+1}$ and $S_{2n-1} \geq S \geq S_{2n} = S_{2n-1} - a_{2n}$. In both cases $|S - S_N| \leq a_{N+1}$. $\blacksquare$

Theorem 3.8b (Cauchy Condensation Test). If $(a_n)$ is a non-negative, decreasing sequence, then $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{k=0}^{\infty} 2^k a_{2^k}$ converges.

Proof. Group the terms of $\sum a_n$. For the lower bound, note:

$a_1 + (a_2 + a_3) + (a_4 + a_5 + a_6 + a_7) + \cdots \geq a_1 + 2a_2 + 4a_4 + 8a_8 + \cdots = \sum_{k=0}^{\infty} 2^k a_{2^k}$

Since each group $(a_{2^k} + \cdots + a_{2^{k+1}-1})$ has $2^k$ terms, each $\geq a_{2^k}$. For the upper Bound:

$a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) + \cdots \leq a_1 + a_2 + 2a_4 + 4a_8 + \cdots \leq a_1 + 2\sum_{k=1}^{\infty} 2^{k-1} a_{2^k}$

If $\sum 2^k a_{2^k}$ converges, the upper bound shows $\sum a_n$ converges. If $\sum a_n$ Converges, the lower bound shows $\sum 2^k a_{2^k}$ converges. $\blacksquare$

Corollary. $\sum_{n=1}^{\infty} 1/n^p$ converges if and only if $p > 1$. Apply the condensation Test: $\sum 2^k \cdot 1/(2^k)^p = \sum 2^{k(1-p)}$A geometric series with ratio $2^{1-p}$ Which converges iff $1 - p \lt 0$I.e., $p > 1$.

Theorem 3.9 (Riemann Rearrangement Theorem). If $\sum a_n$ converges conditionally, then for any $L \in \mathbb{R}$ (or $\pm\infty$), there exists a rearrangement $\sigma : \mathbb{N} \to \mathbb{N}$ such That $\sum_{n=1}^{\infty} a_{\sigma(n)} = L$.

Proof (outline). Let $P = \{n : a_n > 0\}$ and $N = \{n : a_n \lt 0\}$. Since $\sum a_n$ converges Conditionally, both $\sum_{n \in P} a_n = +\infty$ and $\sum_{n \in N} a_n = -\infty$.

To achieve sum $L \in \mathbb{R}$: take positive terms in order until the partial sum exceeds $L$ Then take negative terms until it falls below $L$ Then positive terms again, and so on. Since both The positive and negative subseries diverge, this process can always continue. The terms tend to Zero (since the series converges), so the oscillations around $L$ shrink to zero. $\blacksquare$

Remark. By contrast, every rearrangement of an absolutely convergent series converges to the same sum.

Problem. Determine whether $\sum_{n=1}^{\infty} \frac{n}{2^n}$ converges.

Solution. Apply the ratio test:

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)/2^{n+1}}{n/2^n} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} \lt 1$

By the ratio test, the series converges absolutely. $\blacksquare$

flowchart TD
    A[3_Series] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

A series converges when its partial sums settle down to a finite limit. The challenge is determining this without computing infinitely many terms. Convergence tests are comparison tools: if your series behaves like a known convergent series (geometric, $p$-series), it converges too. The ratio test captures exponential decay of terms — if consecutive terms shrink by a constant factor, the series converges. Conditional convergence is subtle: the alternating harmonic series converges, but rearranging its terms can produce any sum. Absolute convergence is robust: rearrangement preserves the sum. The integral test connects discrete sums to continuous areas, treating the series as a Riemann sum.

Mistake 1: Assuming absolute convergence from conditional convergence A series that converges conditionally (like the alternating harmonic series) does not converge absolutely. Applying the ratio test or root test to a conditionally convergent series gives $L = 1$, which is inconclusive. Always test for absolute convergence separately and use the alternating series test when appropriate.

Mistake 2: Misapplying the ratio test when the limit equals 1 When $\lim |a_{n+1}/a_n| = 1$, the ratio test tells you nothing. Many students conclude convergence or divergence incorrectly in this case. For example, $\sum 1/n$ and $\sum 1/n^2$ both give ratio test limit $1$, yet the first diverges and the second converges. Use another test when $L = 1$.

Mistake 3: Confusing rearrangement of absolutely and conditionally convergent series The Riemann rearrangement theorem states that any conditionally convergent series can be rearranged to converge to any value. Students often assume that rearranging a series preserves its sum, which is only true for absolutely convergent series. Always check absolute convergence before rearranging terms.

• Sequences and Limits: Series convergence is defined through the convergence of partial sums, which are sequences studied in the preceding chapter.
• Sequences and Series of Functions: The convergence tests for numerical series extend to function series through the Weierstrass M-test and uniform convergence.
• Complex Numbers Review: Complex series and power series in the complex plane extend the convergence theory to two dimensions.
• Fourier Series: Fourier series are infinite series of trigonometric functions whose convergence depends on the tests developed in this chapter.
• Quantum Mechanics
• Graph Theory
• Classical Mechanics
• Electromagnetism