Sequences and Series

Last updated: August 2026

Disclaimer: These are my personal notes compiled for my own reference and learning. They may contain errors, incomplete information, or personal interpretations. While I strive for accuracy, these notes are not peer-reviewed and should not be considered authoritative sources. Please consult official textbooks, research papers, or other reliable sources for academic or professional purposes.

Contents

  1. Sequences and the definition of a limit
  2. Completeness, and the Monotone Convergence Theorem
  3. Bolzano–Weierstrass
  4. Cauchy sequences: convergence without knowing the limit
  5. Series as sequences of partial sums
  6. Convergence tests, derived rather than listed
  7. Absolute vs. conditional convergence, and rearrangement
  8. Power series and the radius of convergence
  9. Computation
  10. Common pitfalls
  11. Connections
  12. References

1. Sequences and the definition of a limit

Definition

A sequence $\{a_n\}_{n=1}^\infty$ converges to $L$, written $a_n\to L$, if

$$\forall\epsilon>0\ \exists N\in\mathbb{N} \text{ such that } n\geq N \Rightarrow |a_n-L|<\epsilon.$$

Read this as a two-player commitment: whatever tolerance $\epsilon$ is demanded, the sequence must eventually (past some index $N$, which may depend on $\epsilon$) stay within that tolerance forever after — not just visit it occasionally.

Proposition

Limits are unique, and every convergent sequence is bounded.

Proof. Uniqueness: if $a_n\to L$ and $a_n\to L'$ with $L\neq L'$, take $\epsilon=|L-L'|/2>0$; eventually $|a_n-L|<\epsilon$ and $|a_n-L'|<\epsilon$ simultaneously, so by the triangle inequality $|L-L'|\leq|L-a_n|+|a_n-L'|<2\epsilon=|L-L'|$, a contradiction. Boundedness: taking $\epsilon=1$ in the definition, all but finitely many terms ($n\geq N$) lie in $(L-1,L+1)$; the finitely many terms before $N$ are individually bounded, so the whole sequence is bounded by the max of these finitely many values and $|L|+1$. ∎

2. Completeness, and the Monotone Convergence Theorem

The $\epsilon$-$N$ definition only tells you how to verify a candidate limit — it says nothing about when a limit exists. That existence question is where $\mathbb{R}$'s defining property enters:

Completeness axiom

Every nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound (supremum) in $\mathbb{R}$.

This is not a theorem — it is what distinguishes $\mathbb{R}$ from $\mathbb{Q}$, which fails it ($\{q\in\mathbb{Q}:q^2<2\}$ is bounded above in $\mathbb{Q}$ but has no rational supremum). Every existence result in this note ultimately traces back to this one axiom.

Theorem (Monotone Convergence Theorem)

Every bounded, monotone (increasing or decreasing) sequence converges.

Proof. Suppose $\{a_n\}$ is increasing and bounded above. By completeness, $L=\sup\{a_n : n\in\mathbb{N}\}$ exists. Given $\epsilon>0$, $L-\epsilon$ is not an upper bound (else $L$ would not be the least one), so some $a_N>L-\epsilon$; since the sequence is increasing, $a_n\geq a_N>L-\epsilon$ for all $n\geq N$, and $a_n\leq L$ always (as $L$ is an upper bound), so $|a_n-L|<\epsilon$ for $n\geq N$. The decreasing case is symmetric, using the infimum. ∎

Note precisely what was and was not needed: monotonicity plus boundedness gives existence of a limit without ever exhibiting it — a genuinely different kind of conclusion from computing a limit directly, and one that would be false over $\mathbb{Q}$ (an increasing, bounded sequence of rationals can have an irrational supremum, hence no limit within $\mathbb{Q}$).

3. Bolzano–Weierstrass

Theorem

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

Proof. Every sequence has a monotone subsequence. Call index $n$ a "peak" if $a_n\geq a_m$ for all $m>n$. If there are infinitely many peaks $n_1<n_2<\cdots$, then $\{a_{n_k}\}$ is (weakly) decreasing by definition of peak — a monotone subsequence. If there are only finitely many peaks, pick $n_1$ past the last peak; since $n_1$ is not a peak, some $n_2>n_1$ has $a_{n_2}>a_{n_1}$; since $n_2$ is also not a peak, some $n_3>n_2$ has $a_{n_3}>a_{n_2}$; continuing gives a strictly increasing subsequence. Either way, a monotone subsequence exists. This subsequence is bounded (inheriting the bound on the original sequence), so it converges by the Monotone Convergence Theorem. ∎

This proof is worth pausing on: it produces the monotone subsequence by a purely combinatorial argument about indices (no analysis yet), then hands off to Section 2's theorem for the actual convergence — a clean illustration of how a small number of foundational results (completeness $\Rightarrow$ MCT $\Rightarrow$ Bolzano–Weierstrass) generate the rest of the theory.

4. Cauchy sequences: convergence without knowing the limit

Definition

$\{a_n\}$ is Cauchy if $\forall\epsilon>0\ \exists N$ such that $m,n\geq N \Rightarrow |a_m-a_n|<\epsilon$.

Theorem (Cauchy criterion)

A sequence in $\mathbb{R}$ converges if and only if it is Cauchy.

Proof. ($\Rightarrow$) If $a_n\to L$: given $\epsilon>0$, eventually $|a_n-L|<\epsilon/2$, so for $m,n$ both past that point, $|a_m-a_n|\leq|a_m-L|+|L-a_n|<\epsilon$. ($\Leftarrow$) If $\{a_n\}$ is Cauchy, it is bounded (same argument as boundedness of convergent sequences, using the Cauchy condition with $\epsilon=1$ in place of a known limit), so by Bolzano–Weierstrass it has a convergent subsequence $a_{n_k}\to L$. Given $\epsilon>0$, the Cauchy condition gives $N$ with $|a_m-a_n|<\epsilon/2$ for $m,n\geq N$; choosing $n_k\geq N$ with $|a_{n_k}-L|<\epsilon/2$ gives, for any $n\geq N$, $|a_n-L|\leq|a_n-a_{n_k}|+|a_{n_k}-L|<\epsilon$ — the whole sequence, not just the subsequence, converges to $L$. ∎

The genuine payoff: the Cauchy condition only references the sequence's own terms, so it can certify convergence before knowing the limit — indispensable whenever the limit is exactly the unknown object being constructed (as in the mean-square convergence arguments used to construct the causal MA($\infty$) representation of an AR(1) process in the time series note). And exactly like the Monotone Convergence Theorem, this theorem is specific to complete spaces: $\mathbb{Q}$ has Cauchy sequences (e.g. truncated decimal expansions of $\sqrt2$) that do not converge within $\mathbb{Q}$.

5. Series as sequences of partial sums

Definition

$\sum_{n=1}^\infty a_n$ converges to $S$ if the sequence of partial sums $S_N=\sum_{n=1}^N a_n$ converges to $S$ in the sense of Section 1.

This is not a new theory — every fact about sequences above applies verbatim to $\{S_N\}$. In particular the Cauchy criterion for $\{S_N\}$ reads $|S_m-S_n|=\left|\sum_{k=n+1}^m a_k\right|<\epsilon$ for $m>n\geq N$, the form used repeatedly below.

6. Convergence tests, derived rather than listed

Divergence test

If $\sum a_n$ converges then $a_n\to0$.

Immediate from the Cauchy criterion with $m=n+1$: $|a_{n+1}|<\epsilon$ eventually. The converse is false — the standard counterexample is below.

Geometric series

$\sum_{n=0}^\infty r^n = \dfrac{1}{1-r}$ for $|r|<1$; diverges for $|r|\geq1$.

$S_N=\sum_{n=0}^{N-1}r^n$ satisfies $S_N-rS_N = 1-r^N$ (telescoping), so $S_N=\dfrac{1-r^N}{1-r}$ for $r\neq1$; $r^N\to0$ iff $|r|<1$, giving the limit $\frac{1}{1-r}$. For $|r|\geq1$, $r^N$ does not $\to0$, so $S_N$ does not converge (and diverges to $+\infty$ for $r=1$ trivially).

Integral test

If $f$ is positive, continuous, decreasing on $[1,\infty)$ and $a_n=f(n)$, then $\sum a_n$ converges iff $\int_1^\infty f(x)\,dx$ converges.

Proof. Since $f$ is decreasing, on $[k,k+1]$: $f(k+1)\leq f(x)\leq f(k)$, so integrating gives $f(k+1)\leq\int_k^{k+1}f(x)\,dx\leq f(k)$. Summing $k=1,\ldots,N-1$: $\sum_{k=2}^N a_k \leq \int_1^N f(x)\,dx \leq \sum_{k=1}^{N-1}a_k$. Both the partial sums and $\int_1^N f$ are increasing in $N$ (integrand and terms positive), and each side is bounded iff the other is, by the sandwiching inequality — so by the Monotone Convergence Theorem, one converges iff the other does. ∎

This is precisely what proves the $p$-series result $\sum 1/n^p$ converges iff $p>1$: $\int_1^\infty x^{-p}\,dx = \frac{1}{p-1}$ converges iff $p>1$ (direct computation). In particular $p=1$ (the harmonic series) diverges — while $a_n=1/n\to0$, confirming the divergence test's converse failure claimed above.

Comparison and ratio tests

Comparison: if $0\leq a_n\leq b_n$ eventually and $\sum b_n$ converges, so does $\sum a_n$. Ratio test: if $a_n>0$ and $L=\lim a_{n+1}/a_n$ exists, $\sum a_n$ converges if $L<1$, diverges if $L>1$, and is inconclusive if $L=1$.

Proof (ratio test, via comparison). If $L<1$, pick $r$ with $L<r<1$; eventually (past some $N$) $a_{n+1}/a_n<r$, so $a_{N+k}<a_N r^k$ for $k\geq0$ — comparison against the convergent geometric series $\sum a_Nr^k$ gives convergence. If $L>1$, eventually $a_{n+1}>a_n$, so $a_n\not\to0$, and the divergence test applies. At $L=1$ both $\sum1/n$ (diverges) and $\sum1/n^2$ (converges, by the integral test) have ratio $\to1$, so the test cannot distinguish them — genuinely inconclusive, not merely "harder." ∎
Alternating series test (Leibniz)

If $b_n\geq0$ is decreasing with $b_n\to0$, then $\sum(-1)^{n+1}b_n$ converges, and the error after $N$ terms is at most $b_{N+1}$.

Proof. The even partial sums $S_{2k}$ are increasing ($S_{2k+2}-S_{2k}=b_{2k+1}-b_{2k+2}\geq0$ by the decreasing hypothesis) and bounded above by $b_1$ (regroup $S_{2k}=b_1-(b_2-b_3)-\cdots-b_{2k}\leq b_1$), so by the Monotone Convergence Theorem $S_{2k}\to S$ for some $S$; and $S_{2k+1}-S_{2k}=b_{2k+1}\to0$ forces the odd partial sums to the same limit. Since $S$ lies between any two consecutive partial sums, $|S_N-S|\leq|S_{N+1}-S_N|=b_{N+1}$. ∎

7. Absolute vs. conditional convergence, and rearrangement

Proposition

If $\sum|a_n|$ converges, then $\sum a_n$ converges (absolute convergence implies convergence).

Proof. By the Cauchy criterion applied to $\sum|a_n|$: $\left|\sum_{k=n+1}^m|a_k|\right|<\epsilon$ eventually. By the triangle inequality, $\left|\sum_{k=n+1}^m a_k\right|\leq\sum_{k=n+1}^m|a_k|<\epsilon$ for the same range — so $\{S_N\}$ is itself Cauchy, hence converges. ∎

The alternating harmonic series $\sum(-1)^{n+1}/n$ converges (Leibniz, Section 6) but $\sum1/n$ diverges (harmonic series) — conditional convergence: convergent, but not absolutely. This distinction is not a technicality:

Theorem (Riemann rearrangement, stated without proof)

If $\sum a_n$ converges conditionally, then for any target $T\in\mathbb{R}\cup\{\pm\infty\}$, some reordering of the same terms sums to $T$.

(See Rudin, 1976, Thm. 3.54, for the proof — a direct construction alternating between adding enough positive terms to exceed $T$ and enough negative terms to drop back below it, which terminates infinitely often only because both the positive and negative parts of a conditionally convergent series individually diverge.) Absolutely convergent series have no such pathology — every rearrangement of an absolutely convergent series converges to the same sum, so "the sum of the series," for an absolutely convergent series, is a property of the underlying set of terms, not of the order they happen to be listed in; for a conditionally convergent series it is not.

Plot of partial sums of the alternating harmonic series in standard order converging to ln(2), alongside three rearrangements of the same terms converging to 1, -1, and 0 respectively.
Figure — Same terms, same series, four different sums. All four curves sum exactly the terms $\pm1,\pm1/2,\pm1/3,\ldots$ of the alternating harmonic series — only the order differs. Standard order gives $\ln2\approx0.693$; each rearranged curve greedily adds positive terms until exceeding its target, then negative terms until dropping back below it, and converges to precisely that target. This is only possible because the series is conditionally, not absolutely, convergent — the positive and negative parts $\sum 1/(2k-1)$ and $\sum1/(2k)$ each diverge to $+\infty$ individually, so there is always "enough left" of either sign to push the partial sum toward any target.

8. Power series and the radius of convergence

Theorem (Cauchy–Hadamard)

For $\sum_{n=0}^\infty c_n(x-a)^n$, let $\rho=\limsup_{n\to\infty}|c_n|^{1/n}$ and $R=1/\rho$ (with $R=\infty$ if $\rho=0$, $R=0$ if $\rho=\infty$). The series converges (absolutely) for $|x-a|<R$ and diverges for $|x-a|>R$; behavior exactly at $|x-a|=R$ is not determined by $R$ alone.

This follows from the root test (the same comparison-to-geometric-series idea as the ratio test in Section 6, applied to $|c_n(x-a)^n|^{1/n}=|c_n|^{1/n}|x-a|$): the terms' $n$-th roots tend to $\rho|x-a|$, giving convergence when $\rho|x-a|<1$, i.e. $|x-a|<1/\rho=R$. The boundary genuinely depends on the specific series, not just $R$: for $\sum x^n/n$ ($R=1$), $x=-1$ gives the convergent alternating harmonic series while $x=1$ gives the divergent harmonic series — two different behaviors at the same radius. Within the radius of convergence, a power series can be differentiated and integrated term by term, and the result is again a power series with the same radius $R$ (Rudin, 1976, Thm. 8.1); the resulting Taylor coefficients $c_n=f^{(n)}(a)/n!$ and the Lagrange remainder formula bounding the error of a degree-$N$ truncation are not re-derived here, but are exactly the tool used to justify the Maclaurin series listed in a calculus note's treatment of Taylor series.

Log-scale plot of partial sums of the power series sum (x/3)^n for x=2 (inside the radius, converges), x=3 (exactly at the radius, diverges), and x=4 (outside the radius, diverges faster).
Figure — The Cauchy–Hadamard radius is a genuine dividing line, verified computationally. $\sum(x/3)^n$ has $c_n=(1/3)^n$, so $|c_n|^{1/n}=1/3$ exactly (no limit needed — verified numerically in Section 9) and $R=3$. At $x=2<R$ the partial sums flatten to a finite limit; at $x=4>R$ they diverge geometrically; at $x=3=R$ they still diverge, but only linearly ($(x/3)^n=1$ for every $n$, so $S_N=N+1$) — a qualitatively different, slower kind of divergence than outside the radius, illustrating that "at the boundary" is its own regime, not a blend of the two sides.

9. Computation

The figures above are generated by sequences-series/generate_figures.py. The snippet below verifies the Cauchy–Hadamard radius numerically for the figure's series, and checks the ratio test's convergent case against a series with a known closed form.

import numpy as np

# Cauchy-Hadamard radius for c_n = (1/3)^n
n = np.arange(1, 200)
c_n = (1.0 / 3.0) ** n
root_n = c_n ** (1.0 / n)
print(f"|c_n|^(1/n) at n=199: {root_n[-1]:.6f}  ->  R = 1/that = {1/root_n[-1]:.6f}  (exact R = 3)")

# Ratio test convergent case: sum n^2 / 2^n, closed form x(1+x)/(1-x)^3 at x=1/2
def partial_sum(N):
    return sum(k**2 / 2**k for k in range(1, N + 1))

for N in (10, 20, 40):
    print(f"N={N:3d}: partial sum = {partial_sum(N):.6f}")
x = 0.5
print("closed form x(1+x)/(1-x)^3 at x=1/2:", x * (1 + x) / (1 - x)**3)

Actual output:

|c_n|^(1/n) at n=199: 0.333333  ->  R = 1/that = 3.000000  (exact R = 3)
N= 10: partial sum = 5.857422
N= 20: partial sum = 5.999537
N= 40: partial sum = 6.000000
closed form x(1+x)/(1-x)^3 at x=1/2: 6.0

The estimated radius matches $R=3$ to six decimal places, and the ratio-test-justified series visibly converges toward its exact closed-form value $6$ as $N$ grows, consistent with Section 6's proof rather than merely "looking like it settles down."

10. Common pitfalls

Pitfall — $a_n\to0$ does not imply $\sum a_n$ converges

The divergence test is a one-directional necessary condition, not sufficient. The harmonic series $\sum1/n$ is the standard counterexample: terms $\to0$, series diverges (Section 6).

Pitfall — Ratio/root test at $L=1$ is not "probably converges" or "probably diverges"

It is genuinely uninformative — $\sum1/n$ and $\sum1/n^2$ both have ratio limit $1$ and opposite convergence behavior (Section 6). A different test (often the integral test or a more refined comparison) is required, not a judgment call about how close to $1$ the ratio is.

Pitfall — Rearranging a conditionally convergent series changes its sum

Not a pathological edge case reserved for pure mathematicians: any computation that reorders terms of a conditionally convergent sum (e.g. summing floating-point contributions in a different order, or reindexing a double sum without justifying absolute convergence first) can silently change the numerical answer, exactly as demonstrated in the figure above. Absolute convergence is the standing assumption that makes term reordering safe (Section 7); check it before reordering, not after noticing a discrepancy.

Pitfall — The radius of convergence says nothing about the boundary

$|x-a|=R$ requires separate case-by-case analysis; different power series with the same $R$ can converge at both boundary points, one, or neither (Section 8's example already shows two different behaviors at the two endpoints of a single series).

11. Connections

12. References