Infinite series: convergence and the standard tests

Intuition [Beginner]

An infinite series is what you get when you keep adding numbers from a list that never ends. Start with the list $1/2$, $1/4$, $1/8$, $1/16$, and so on — each term is half the one before. Track the running total: $1/2$, then $3/4$, then $7/8$, then $15/16$. The running totals approach $1$ but never overshoot it. The series converges to $1$.

Not every infinite addition settles down. The list $1$, $2$, $3$, $4$, and so on gives running totals $1$, $3$, $6$, $10$, growing without bound. This series diverges. Even the list $1$, $1/2$, $1/3$, $1/4$, and so on — whose terms shrink — gives a running total that grows without bound, just very slowly.

The central question: given a list of numbers being added up forever, does the running total approach a finite target, or does it escape? The convergence tests in this unit are systematic rules for answering that question without having to compute the running total forever.

This concept exists because calculus is built on adding up infinitely many pieces — areas under curves, accumulated change, solutions to equations — and every such computation reduces to deciding whether an infinite series converges.

Visual [Beginner]

The picture shows a number line from $0$ to $1$. Above the line, tick marks appear at the partial sums $1/2$, $3/4$, $7/8$, $15/16$, each one closer to $1$ than the last. The distance from each tick to $1$ halves at every step. A second number line below shows the partial sums of $1+1/2+1/3+1/4+\cdots$ marching rightward past every integer, illustrating divergence.

The top number line is the geometric series, the prototypical convergent series. The bottom is the harmonic series, the prototypical divergent series whose terms shrink but not fast enough.

Worked example [Beginner]

Take the series $1/3+1/9+1/27+1/81+\cdots$ where each term is one third of the one before. Compute the running totals:

$s_1=1/3$. Distance to $1/2$ is $1/6$.

$s_2=1/3+1/9=4/9$. Distance to $1/2$ is $1/18$.

$s_3=4/9+1/27=13/27$. Distance to $1/2$ is $1/54$.

Each running total has the form $s_n=(1-(1/3{)}^n)/2$. As $n$ grows, $(1/3{)}^n$ shrinks toward $0$, so $s_n$ approaches $1/2$. The series $1/3+1/9+1/27+\cdots$ converges to $1/2$.

What this tells us: any geometric series $a+ar+a{r}^2+a{r}^3+\cdots$ with ratio $r$ between $-1$ and $1$ converges, because each new term is a smaller correction and the running total settles at $a/(1-r)$.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Fix the real numbers with absolute value from 02.02.01. An infinite series of real numbers is an expression ${\sum }_{n=1}^{\infty }{a}_n$ determined by a sequence $(a_n{)}_{n\ge 1}$ in $\mathbb{R}$. The partial sums are $s_N={\sum }_{n=1}^N{a}_n=a_1+a_2+\cdots +a_N$. The sequence $(s_N{)}_{N\ge 1}$ of partial sums is what gives the series its meaning.

Definition (convergence of a series). The series ${\sum }_{n=1}^{\infty }{a}_n$ converges to $L\in \mathbb{R}$ iff the sequence of partial sums $s_N\to L$ as $N\to \infty$ (in the sense of 02.03.02). Write ${\sum }_{n=1}^{\infty }{a}_n=L$. The series diverges iff the sequence of partial sums does not converge.

Definition (absolute convergence). The series ${\sum }_{n=1}^{\infty }{a}_n$ converges absolutely iff ${\sum }_{n=1}^{\infty }|a_n|$ converges. A series that converges but does not converge absolutely is called conditionally convergent.

Absolute convergence implies convergence (proved as Corollary below in the Key theorem section). The converse fails: the alternating harmonic series ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}/n=1-1/2+1/3-1/4+\cdots$ converges (to $\ln 2$) but does not converge absolutely, because ${\sum }_{n=1}^{\infty }1/n$ diverges.

The presentation follows Apostol ^{[Apostol Ch. 10]} and Rudin ^{[Rudin Ch. 3]}.

Counterexamples to common slips

• Terms shrinking to zero does not guarantee convergence. The harmonic series ${\sum }_{n=1}^{\infty }1/n$ has $a_n=1/n\to 0$ but diverges, as shown in 02.03.02.
• Convergence does not imply absolute convergence. The alternating harmonic series converges conditionally, not absolutely. The distinction matters: absolutely convergent series can be rearranged freely without changing the sum; conditionally convergent series cannot.
• Convergence of $\sum a_n$ does not imply convergence of $\sum a_n^2$. Take $a_n=(-1{)}^n/\sqrt{n}$: the series converges by the alternating series test, but $\sum a_n^2=\sum 1/n$ diverges.
• A convergent series with positive terms can converge to a value smaller than the first term. The geometric series $1/4+1/16+1/64+\cdots$ converges to $1/3$, which is less than the first term $1/4$.

Key theorem with proof [Intermediate+]

Theorem (Comparison test). Let $0\le a_n\le b_n$ for all $n\ge 1$. If ${\sum }_{n=1}^{\infty }{b}_n$ converges, then ${\sum }_{n=1}^{\infty }{a}_n$ converges. Equivalently, if ${\sum }_{n=1}^{\infty }{a}_n$ diverges, then ${\sum }_{n=1}^{\infty }{b}_n$ diverges.

Proof. The partial sums $s_N={\sum }_{n=1}^N{a}_n$ form a monotone increasing sequence (each $a_n\ge 0$), bounded above by the sum ${\sum }_{n=1}^{\infty }{b}_n$:

$$s_N=\sum _{n=1}^N{a}_n\le \sum _{n=1}^N{b}_n\le \sum _{n=1}^{\infty }{b}_n.$$

A monotone increasing bounded-above sequence converges by the monotone convergence theorem, which is a consequence of the Cauchy criterion proved in 02.03.02. $\square$

Corollary (Absolute convergence implies convergence). If ${\sum }_{n=1}^{\infty }|a_n|$ converges, then ${\sum }_{n=1}^{\infty }{a}_n$ converges.

Proof. For any indices $n\ge m$, the triangle inequality gives

$$\left|\sum _{k=m}^n{a}_k\right|\le \sum _{k=m}^n|a_k|.$$

Since $\sum |a_n|$ converges, its partial sums form a Cauchy sequence (by the Cauchy criterion of 02.03.02): for every $\epsilon >0$, some $N$ exists with ${\sum }_{k=m}^n|a_k|<\epsilon$ for $n\ge m\ge N$. The displayed inequality then gives $|{\sum }_{k=m}^n{a}_k|<\epsilon$, so the partial sums of $\sum a_n$ form a Cauchy sequence, and converge. $\square$

Theorem (Ratio test). Let ${\sum }_{n=1}^{\infty }{a}_n$ be a series with $a_n\ne 0$ for all $n$, and set $L={\mathrm{limsup}}_{n\to \infty }|a_{n+1}/a_n|$.

(i) If $L<1$, then ${\sum }_{n=1}^{\infty }{a}_n$ converges absolutely.

(ii) If $L>1$ (including $L=\infty$), then ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

(iii) If $L=1$, the test is inconclusive.

Proof. (i) Suppose $L<1$. Choose $r$ with $L<r<1$. By definition of $\mathrm{limsup}$, only finitely many indices $n$ satisfy $|a_{n+1}/a_n|>r$. So there exists $N_0$ such that $|a_{n+1}/a_n|\le r$ for all $n\ge N_0$. Iterating, $|a_n|\le |a_{N_0}|\cdot r^{n-N_0}$ for $n\ge N_0$. The series ${\sum }_{n=N_0}^{\infty }|a_{N_0}|r^{n-N_0}=|a_{N_0}|/(1-r)$ converges (geometric series with ratio $r<1$). By the comparison test, ${\sum }_{n=N_0}^{\infty }|a_n|$ converges, and adding finitely many terms preserves convergence.

(ii) If $L>1$, then $|a_{n+1}|>|a_n|$ for infinitely many $n$, so $a_n$ does not converge to $0$. A necessary condition for convergence of $\sum a_n$ is $a_n\to 0$; the contrapositive gives divergence.

(iii) For the $p$-series ${\sum }_{n=1}^{\infty }1/n^p$: the ratio is $|a_{n+1}/a_n|=(n/(n+1){)}^p\to 1$ for every $p>0$. The series converges for $p>1$ and diverges for $p\le 1$, so $L=1$ gives no information. $\square$

Theorem (Alternating series test). Let $(a_n)$ be a decreasing sequence of non-negative real numbers with $a_n\to 0$. Then the alternating series ${\sum }_{n=1}^{\infty }(-1{)}^{n+1}{a}_n$ converges. Moreover, the remainder after $N$ terms satisfies $|s-s_N|\le a_{N+1}$.

Proof. Group the partial sums in pairs. For even indices $N=2m$:

$$s_{2m}=(a_1-a_2)+(a_3-a_4)+\cdots +(a_{2m-1}-a_{2m}).$$

Since $a_n$ is decreasing, each grouped term $a_{2k-1}-a_{2k}\ge 0$, so $(s_{2m})$ is monotone increasing. For odd indices $N=2m+1$:

$$s_{2m+1}=a_1-(a_2-a_3)-(a_4-a_5)-\cdots -(a_{2m}-a_{2m+1}).$$

Each grouped term is subtracted, so $(s_{2m+1})$ is monotone decreasing. Both sequences lie in the interval $[s_2,s_1]$ and the difference satisfies

$$s_{2m+1}-s_{2m}=a_{2m+1}\to 0.$$

By the Cauchy criterion from 02.03.02, the full sequence $(s_N)$ converges. The remainder bound: $s_{2m}\le s\le s_{2m+1}$, so $|s-s_{2m}|\le s_{2m+1}-s_{2m}=a_{2m+1}$, and similarly for odd indices. $\square$

Bridge. The comparison test is the foundational reason that absolute convergence reduces to non-negative comparison: a series converges absolutely iff its termwise absolute values are dominated by a convergent non-negative series. This is exactly the reduction that makes the ratio test work — the ratio test compares $|a_n|$ to a geometric series via the comparison test. The alternating series test is dual to the comparison test in the sense that it exploits cancellation rather than dominance, and the central insight is that decreasing terms heading to zero guarantee enough cancellation to produce convergence. Putting these together, the three tests cover the main cases: positive-term series (comparison and ratio), sign-alternating series (alternating series test), and general series (absolute convergence as a reduction to the positive case). The ratio test builds toward the root test and Raabe's test in the Master tier below, where the convergence question is refined for series whose ratio test is inconclusive.

Exercises [Intermediate+]

Advanced results [Master]

The comparison test and ratio test are the entry point to a hierarchy of convergence criteria, each refining the previous one for the boundary case where the simpler test is inconclusive.

Theorem (Root test, Cauchy). Let ${\sum }_{n=1}^{\infty }{a}_n$ be a series of real numbers and set $\alpha ={\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}$.

(i) If $\alpha <1$, the series converges absolutely.

(ii) If $\alpha >1$, the series diverges.

(iii) If $\alpha =1$, the test is inconclusive.

The root test is strictly stronger than the ratio test in the sense that whenever the ratio test gives a conclusion, so does the root test, but not conversely. The inequality $\mathrm{limsup}\sqrt[n]{|a_n|}\le \mathrm{limsup}|a_{n+1}/a_n|$ guarantees that whenever $L<1$ in the ratio test, $\alpha <1$ in the root test. The root test handles series such as $a_n=2^{n+(-1{)}^n}$ where the ratios oscillate between $1/2$ and $8$ but $\sqrt[n]{|a_n|}\to 2>1$ gives divergence. The root test is inconclusive for the same family as the ratio test: $p$-series, where $\sqrt[n]{1/n^p}\to 1$ for all $p$. ^{[Cauchy 1821]}

Theorem (Raabe's test). Let $a_n>0$ for all $n$ and suppose ${\lim }_{n\to \infty }n(a_n/a_{n+1}-1)=\beta$ exists.

(i) If $\beta >1$, then $\sum a_n$ converges.

(ii) If $\beta <1$, then $\sum a_n$ diverges.

(iii) If $\beta =1$, the test is inconclusive.

Raabe's test refines the ratio test at $L=1$. When $|a_{n+1}/a_n|=1-\beta /n+o(1/n)$, the ratio test is inconclusive ($L=1$), but Raabe's test resolves the question: $\beta >1$ gives convergence, $\beta <1$ gives divergence. The boundary is sharp: for $a_n=1/(n{\ln }^{2}n)$, the ratio test gives $L=1$, Raabe's test gives $\beta =1$, and yet the series converges (by the integral test). The test originates with Raabe 1832 Crelle's J. 9.

Theorem (Dirichlet's test). Let $(a_n)$ be a sequence of real numbers with partial sums $A_N={\sum }_{n=1}^N{a}_n$ bounded: $|A_N|\le M$ for all $N$. Let $(b_n)$ be a decreasing sequence of non-negative real numbers with $b_n\to 0$. Then ${\sum }_{n=1}^{\infty }{a}_n{b}_n$ converges.

Dirichlet's test ^{[Dirichlet 1837]} generalises the alternating series test. The alternating series test is the special case $a_n=(-1{)}^{n+1}$ (partial sums bounded by $1$) and $b_n$ decreasing to $0$. The proof uses Abel summation (summation by parts): ${\sum }_{k=1}^N{a}_k{b}_k=A_N{b}_N+{\sum }_{k=1}^{N-1}{A}_k(b_k-b_{k+1})$. Boundedness of $(A_N)$ and $b_n\searrow 0$ force both terms on the right to converge. Dirichlet's test is the workhorse behind the convergence of ${\sum }_{n=1}^{\infty }(\sin n\theta )/n$ for real $\theta$.

Theorem (Abel's test). Let ${\sum }_{n=1}^{\infty }{a}_n$ converge. Let $(b_n)$ be a monotone convergent sequence. Then ${\sum }_{n=1}^{\infty }{a}_n{b}_n$ converges.

Abel's test ^{[Abel 1826]} relaxes Dirichlet's condition on $(a_n)$ from bounded partial sums to convergent partial sums, and compensates by requiring $(b_n)$ to converge (not merely tend to $0$). The proof is the same summation-by-parts calculation, with the extra observation that $A_N{b}_N\to Ab$ where $A=\sum a_n$ and $b=\lim b_n$. Abel's test is the natural tool for series of the form $\sum a_n{r}^n$ where $\sum a_n$ converges and $r\in (0,1]$: the series converges for all such $r$, giving continuity of the power-series function up to the boundary.

Theorem (Riemann rearrangement theorem). If ${\sum }_{n=1}^{\infty }{a}_n$ converges conditionally, then for every real number $L$ there exists a permutation $\sigma :\mathbb{N}\to \mathbb{N}$ such that ${\sum }_{n=1}^{\infty }{a}_{\sigma (n)}=L$. Moreover, there exist permutations for which the rearranged series diverges to $+\infty$, diverges to $-\infty$, or fails to converge to any extended real value.

The theorem ^{[Riemann 1854]} says that conditional convergence is fragile: the sum depends on the order of summation. The proof exploits the positive and negative tails. Set $p_n=\max (a_n,0)$ and $q_n=-\min (a_n,0)$. Conditional convergence forces $\sum p_n=+\infty$ and $\sum q_n=+\infty$ while $p_n\to 0$ and $q_n\to 0$. To hit a target $L$: add positive terms until the partial sum exceeds $L$, then add negative terms until it undershoots, then repeat. Because both tails are divergent, each crossing achieves a tighter overshoot, and because the terms tend to $0$, the oscillations shrink to $L$. Absolute convergence, by contrast, is permutation-invariant: every rearrangement of an absolutely convergent series converges to the same sum.

Synthesis. The convergence tests form a hierarchy of increasing specificity, and the foundational reason is that each test refines how the $n$-th term is compared to a geometric decay. The ratio test compares $|a_{n+1}/a_n|$ to a constant $r<1$; the root test compares $\sqrt[n]{|a_n|}$ to $1$ via $\mathrm{limsup}$; Raabe's test compares the ratio to $1-\beta /n$, extracting one more term in the asymptotic expansion. This is exactly the pattern by which convergence criteria generalise: each test in the chain handles the boundary case the previous test could not resolve. The central insight is that absolute convergence identifies a series with a summable family indexed by $\mathbb{N}$, for which order is irrelevant; conditional convergence is the residual case where cancellation between positive and negative terms produces a finite sum, but the cancellation is order-dependent. Putting these together with the Riemann rearrangement theorem gives the complete picture: the bridge is between the algebra of unordered summation (absolute convergence, permutation-invariance) and the analytic fragility of ordered summation (conditional convergence, rearrangement-sensitive). The pattern recurs in double series, where Fubini's theorem requires absolute convergence for the interchange of summation order, and in Fourier series, where pointwise convergence of conditionally convergent trigonometric series demands careful analysis.

Full proof set [Master]

Proposition (Root test). Let $\alpha ={\mathrm{limsup}}_{n\to \infty }\sqrt[n]{|a_n|}$. If $\alpha <1$ then $\sum a_n$ converges absolutely. If $\alpha >1$ then $\sum a_n$ diverges.

Proof. Suppose $\alpha <1$. Choose $r$ with $\alpha <r<1$. By definition of $\mathrm{limsup}$, the set $\{n:\sqrt[n]{|a_n|}>r\}$ is finite. So for all $n\ge N_0$ (some index $N_0$), $|a_n|\le r^n$. The geometric series $\sum r^n$ converges ($r<1$), and the comparison test gives $\sum |a_n|<\infty$.

Suppose $\alpha >1$. Then $\sqrt[n]{|a_n|}>1$ for infinitely many $n$, so $|a_n|>1$ for infinitely many $n$, and $a_n\nrightarrow 0$. A series whose terms do not converge to zero diverges. $\square$

Proposition (Raabe's test, convergence case). If $a_n>0$ and $n(a_n/a_{n+1}-1)\to \beta >1$, then $\sum a_n$ converges.

Proof. Choose $\epsilon >0$ with $\beta -\epsilon >1$. There exists $N_0$ such that for $n\ge N_0$:

$$\frac{a_n}{a_{n+1}}>1+\frac{\beta -\epsilon }{n}.$$

Equivalently, $a_{n+1}<a_n\cdot n/(n+\beta -\epsilon )$. Iterating from index $N_0$ to $N_0+k-1$:

$$a_{N_0+k}<a_{N_0}\prod _{j=0}^{k-1}\frac{N_0+j}{N_0+j+\beta -\epsilon }.$$

Taking logarithms and using $\ln (1-x)\le -x$ for $0<x<1$:

$$\ln a_{N_0+k}\le \ln a_{N_0}-(\beta -\epsilon )\sum _{j=0}^{k-1}\frac{1}{N_0+j}\le \ln a_{N_0}-(\beta -\epsilon ){\int }_{N_0}^{N_0+k}\frac{dx}{x}.$$

The integral evaluates to $\ln ((N_0+k)/N_0)$, giving $a_{N_0+k}\le a_{N_0}\cdot N_0^{\beta -\epsilon }/(N_0+k{)}^{\beta -\epsilon }$. Since $\beta -\epsilon >1$, the series $\sum 1/n^{\beta -\epsilon }$ converges, and by comparison $\sum a_n$ converges. $\square$

Proposition (Dirichlet's test). If $|A_N|\le M$ for all $N$ and $b_n\searrow 0$, then $\sum a_n{b}_n$ converges.

Proof. By Abel summation, for $N\ge 2$:

$$\sum _{k=1}^N{a}_k{b}_k=A_N{b}_N+\sum _{k=1}^{N-1}{A}_k(b_k-b_{k+1}).$$

Since $b_N\to 0$ and $|A_N|\le M$, the first term satisfies $|A_N{b}_N|\le M{b}_N\to 0$. For the second term, $b_k-b_{k+1}\ge 0$ (because $b_n$ is decreasing), so

$$\sum _{k=1}^{N-1}|A_k(b_k-b_{k+1})|\le M\sum _{k=1}^{N-1}(b_k-b_{k+1})=M(b_1-b_N)\le M{b}_1.$$

The partial sums of $\sum A_k(b_k-b_{k+1})$ form a bounded monotone sequence (each term has fixed sign), hence converge. Adding $A_N{b}_N\to 0$, the series $\sum a_n{b}_n$ converges. $\square$

Proposition (Abel's test). If $\sum a_n$ converges and $(b_n)$ is monotone convergent, then $\sum a_n{b}_n$ converges.

Proof. Set $A={\sum }_{n=1}^{\infty }{a}_n$ and $b=\lim b_n$. Define ${\tilde{b}}_n=b_n-b$. Then ${\tilde{b}}_n$ is monotone with ${\tilde{b}}_n\to 0$. The partial sums of $a_n$ converge to $A$, hence are bounded. By Dirichlet's test applied to $a_n$ and ${\tilde{b}}_n$, the series $\sum a_n(b_n-b)$ converges. Since $\sum a_{n}b=b\cdot A$ also converges, $\sum a_n{b}_n=\sum a_n(b_n-b)+\sum a_{n}b$ converges. $\square$

Proposition (Riemann rearrangement theorem). If $\sum a_n$ converges conditionally, then for any $L\in \mathbb{R}$ there exists a permutation $\sigma$ with $\sum a_{\sigma (n)}=L$.

Proof. Define $p_n=\max (a_n,0)$ and $q_n=\max (-a_n,0)$. Then $a_n=p_n-q_n$, $p_n\ge 0$, $q_n\ge 0$. Since $\sum a_n$ converges conditionally, both $\sum p_n=+\infty$ and $\sum q_n=+\infty$ (if either converged, $\sum |a_n|$ would converge), and $p_n\to 0$, $q_n\to 0$.

Fix $L\in \mathbb{R}$. Construct the permutation inductively. Let $S_0=0$. At stage $k$, if $S_{k-1}\le L$, take the next unused positive term $p_{n_k}$ and set $S_k=S_{k-1}+p_{n_k}$. If $S_{k-1}>L$, take the next unused negative term $-q_{m_k}$ and set $S_k=S_{k-1}-q_{m_k}$. Because $\sum p_n=+\infty$, every time the partial sum dips below $L$ it eventually overshoots; because $\sum q_n=+\infty$, every time it overshoots it eventually dips below. The terms used satisfy $p_{n_k}\to 0$ and $q_{m_k}\to 0$, so $|S_k-L|\to 0$. Every index is eventually consumed because each step uses one unused index. The rearranged series converges to $L$. $\square$

Connections [Master]

• Cauchy sequences and Bolzano-Weierstrass 02.03.02. The definition of series convergence is the convergence of the partial-sum sequence, and every convergence test for series ultimately rests on the Cauchy criterion and the completeness of $\mathbb{R}$ proved in that unit. The comparison test invokes the monotone convergence theorem, which is one of the four equivalent faces of completeness from 02.03.02.

• Metric space 02.01.05. The Cauchy criterion for series — a series converges iff its partial sums form a Cauchy sequence — is the metric-space Cauchy criterion applied to $\mathbb{R}$ with the Euclidean metric. The absolute-convergence theory generalises to Banach spaces and normed vector spaces, where the comparison test and ratio test carry over verbatim.

• Banach spaces 02.11.04. Absolute convergence characterises completeness in normed vector spaces: a normed space is a Banach space (complete) iff every absolutely convergent series converges. This equivalence, proved in 02.11.04, identifies absolute convergence as the series-theoretic face of the completeness axiom. The Riemann rearrangement theorem is the negative statement: without absolute convergence, the sum depends on the ordering, so completeness alone does not rescue conditional convergence from rearrangement fragility.

Historical & philosophical context [Master]

Augustin-Louis Cauchy's Cours d'analyse of 1821 ^{[Cauchy 1821]} established the first systematic convergence criteria for infinite series, including the root test and the ratio test. Cauchy worked in the setting of real numbers as magnitudes, without the axiomatisation that came later. His criterion for convergence — the partial sums form a Cauchy sequence — converted the informal "adding up infinitely many terms" into a precise $\epsilon$-statement, and the root and ratio tests gave the first practical decision procedures.

Peter Gustav Lejeune Dirichlet's 1837 paper on the convergence of Fourier series ^{[Dirichlet 1837]} introduced the summation-by-parts technique and what is now called Dirichlet's test, in the context of proving that $\sum (\sin n\theta )/n$ converges for every real $\theta$. Niels Henrik Abel's 1826 Recherches sur les fonctions ^{[Abel 1826]} proved what is now called Abel's test as a tool for studying binomial series and power series on the boundary of their domain of convergence. Bernhard Riemann's 1854 Habilitationsschrift ^{[Riemann 1854]}, published posthumously in 1867, contained the rearrangement theorem as part of his foundational work on the representability of functions by trigonometric series. Riemann showed that conditionally convergent series could be rearranged to converge to any prescribed real number, establishing absolute convergence as the correct notion for permutation-invariant summation.

Bibliography [Master]

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}

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}

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}

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}