Series | Mathematics - Wyatt's Notes

Attempt these from memory before reading.

• What does it mean for $\sum a_n$ to converge?
• State the ratio test. When is it inconclusive?
• What is the difference between absolute and conditional convergence?
• Does $\sum 1/n$ converge? Does $\sum (-1)^n/n$?
• Is the sum of a conditionally convergent series independent of the order of the terms?

Check your answers against the text. The last prompt is the one most people answer wrongly, and the answer is the most surprising result on this page.

• Does $\sum_{n=1}^{\infty} 1/n$ converge?
• Does $\sum_{n=1}^{\infty} 1/n^2$ converge?
• Does $\sum_{n=1}^{\infty} (-1)^n/n$ converge?

The first diverges. The second converges to $\pi^2/6$. The third converges to $\ln 2$ — and if you rearrange its terms you can make it converge to any real number you like.

Three series whose terms all tend to zero, with three different behaviours. The whole of this page is the machinery for telling them apart, and the machinery matters: a series is the discrete analogue of an integral, power series are how functions are computed, and Fourier series are how signals are analysed. Getting the convergence tests right is the price of entry to all of that.

The paradox to hold onto. Adding infinitely many positive terms can give a finite answer ($1/2 + 1/4 + 1/8 + \cdots = 1$), and adding infinitely many terms that tend to zero can give an infinite answer ($1 + 1/2 + 1/3 + \cdots$). “Tends to zero” is necessary and not sufficient, and the reason is the accumulated total, not the individual terms.

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:

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

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:

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:

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:

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:

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

The Cauchy condensation test is the underused one, and it is the fastest way to settle the $p$-series. It says $\sum a_n$ (positive, decreasing) converges exactly when $\sum 2^k a_{2^k}$ does. Applying it to $\sum 1/n^p$ gives $\sum 2^k / 2^{kp} = \sum 2^{k(1-p)}$, a geometric series in $2^{1-p}$ — which converges exactly when $p > 1$. One line, no integral comparison.

Take the alternating harmonic series $\sum (-1)^{n+1}/n = \ln 2 \approx 0.693$. Its positive terms are $1, 1/3, 1/5, \dots$ with divergent sum, and its negative terms are $-1/2, -1/4, -1/6, \dots$ with divergent sum.

To make it sum to $1.5$ instead: take positive terms until the partial sum exceeds $1.5$ ($1 + 1/3 + 1/5 = 1.533$), then negative terms until it drops below ($-1/2$ gives $1.033$), then positive again, and so on. The oscillations shrink because the terms tend to zero, so the rearranged series converges — to $1.5$ rather than $\ln 2$.

Why this matters beyond the puzzle. Conditional convergence means the sum depends on the order, so it is not a property of the terms at all. Any computation involving such a series must fix an order. Numerically, summing a conditionally convergent series in floating point gives order-dependent answers, which is a real source of error in signal processing and numerical PDE.

Absolutely convergent series have no such problem: every rearrangement converges to the same sum, because the positive and negative parts each converge.

Exercise. Show that the positive terms of a conditionally convergent series have divergent sum, and likewise the negative terms. (Hint: if both converged, the series would converge absolutely.)

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

• Continuity
• Differentiability
• Problem Set
• Riemann Integration