Summation by Parts, Uniform Cancellation, and Abel's Endpoint Theorem

Learning goals and dependencies

Absolute convergence is sufficient for many operations, but it does not describe every convergent series. Cancellation can keep long sums small even when the absolute values have a divergent sum. This chapter derives a finite summation identity and uses it to prove numerical and uniform Dirichlet tests. It then proves Abel's endpoint theorem, including the uniform convergence that justifies continuity at a convergent power-series endpoint.

We use the Cauchy criterion, uniform convergence and its preservation of continuity, monotone real sequences, elementary trigonometric identities, and the geometric series. When logarithms enter a worked example, the fundamental theorem of calculus is also used. The trigonometric construction already supplies the addition formulas; no Fourier-series theory is presupposed.

1. A finite identity before an infinite conclusion

Let $m\le N$ be integers. For real sequences $a_n,b_n$, put

$$B_{m,k}=\sum_{n=m}^k b_n\quad(k\ge m),\qquad B_{m,m-1}=0.$$

Because $b_n=B_{m,n}-B_{m,n-1}$, subtracting two finite sums gives

$$\begin{aligned} \sum_{n=m}^N a_nb_n &=\sum_{n=m}^N a_nB_{m,n}-\sum_{n=m}^N a_nB_{m,n-1}\\ &=a_NB_{m,N}+\sum_{n=m}^{N-1}(a_n-a_{n+1})B_{m,n}. \end{aligned}$$

In the second line, the zero term at $n=m$ in the second sum disappears, and shifting its remaining indices aligns $a_{n+1}B_{m,n}$ with the first sum. This is summation by parts. It is an exact finite identity and has no convergence hypothesis.

If $a_n\ge0$ is nonincreasing and $|B_{m,k}|\le B$ throughout the indicated block, the triangle inequality gives

$$\left|\sum_{n=m}^N a_nb_n\right| \le Ba_N+B\sum_{n=m}^{N-1}(a_n-a_{n+1})=Ba_m.$$

The differences are nonnegative, and their sum is $a_m-a_N$. This telescoping step is exactly where monotonicity is used. A bound on the individual $b_n$ would not replace a bound on their sums.

2. Dirichlet's test for numerical series

Theorem 1. Suppose the partial sums $P_N=\sum_{n=1}^N b_n$ are bounded, say $|P_N|\le M$, and $a_n\ge0$ is nonincreasing and tends to zero. Then $\sum a_nb_n$ converges. Moreover, if its sum is $s$,

$$\left|s-\sum_{n=1}^N a_nb_n\right|\le2Ma_{N+1}.$$

Proof. Set $P_0=0$. Every block sum satisfies

$$|B_{m,k}|=|P_k-P_{m-1}|\le2M.$$

The finite identity therefore gives $|\sum_{n=m}^N a_nb_n|\le2Ma_m$. Given $\varepsilon>0$, choose $m_0$ so that $2Ma_m<\varepsilon$ for all $m\ge m_0$; if $M=0$, every $b_n=P_n-P_{n-1}$ is zero and the conclusion follows directly. The tail estimate is the Cauchy criterion, proving convergence in $\mathbb R$. For the remainder after $N$, apply the estimate with starting index $N+1$ and let the finite upper index tend to infinity. Absolute values are continuous, so the inequality is preserved in that limit. $\square$

When a sharper direct block bound $|B_{m,k}|\le B$ is known, the same proof yields remainder at most $Ba_{N+1}$. For alternating coefficients $b_n=(-1)^{n-1}$, every block sum has magnitude at most one. Thus this argument includes the usual alternating-series remainder bound $a_{N+1}$.

Periodic coefficients and their mean

Suppose $b_n$ repeats a block $\beta_1,\ldots,\beta_p$ whose sum is zero. Every complete period contributes zero. An initial partial sum is therefore one of the finitely many values $0,\beta_1,\beta_1+\beta_2,\ldots,\beta_1+\cdots+\beta_{p-1}$, and is bounded. Dirichlet's test applies to any nonnegative, nonincreasing weight tending to zero.

If the period has nonzero mean $\mu=p^{-1}\sum\beta_j$, write $b_n=\mu+d_n$. The new periodic coefficients $d_n$ have zero mean and bounded partial sums. For weights $1/n$,

$$\sum_{n=1}^N\frac{b_n}{n}=\mu\sum_{n=1}^N\frac1n+\sum_{n=1}^N\frac{d_n}{n}.$$

The second term has a finite limit, while the first tends to $+\infty$ if $\mu>0$ and to $-\infty$ if $\mu<0$. Thus periodicity by itself is not the cancellation hypothesis; zero mean supplies it.

Worked example 1: three-term cancellation, not alternating signs. Let the coefficients repeat $2,-1,-1$, and consider

$$s=\sum_{n=1}^{\infty}\frac{b_n}{n+2}.$$

The initial partial sums take values only $0,1,2$, so any difference of two such sums has magnitude at most two. The direct block bound is $B=2$. Since $a_n=1/(n+2)$ decreases to zero, the series converges and its remainder after $N$ terms is at most $2/(N+3)$. The absolute series diverges, because $|b_n|\ge1$ and hence $|b_n|/(n+2)\ge1/(n+2)$.

The first six terms give

$$s_6=\frac23-\frac14-\frac15+\frac26-\frac17-\frac18=\frac{79}{280}.$$

Consequently $|s-79/280|\le2/9$. This is a deliberately coarse but valid certificate. It identifies a finite, reproducible approximation and an analytic error bound without pretending that the approximation is the exact sum.

3. One bound for every parameter: uniform Dirichlet

A parameter-dependent series may converge at every parameter value while having no common rate of convergence. The uniform Cauchy criterion requires that for each $\varepsilon>0$ one starting index controls every parameter and every later tail.

Theorem 2. Let $E$ be a set. Suppose, for every $x\in E$,

1. $a_n(x)\ge0$ and $a_{n+1}(x)\le a_n(x)$;
2. $\sup_{x\in E}a_n(x)\to0$;
3. a finite constant $B$ satisfies $|\sum_{n=m}^N b_n(x)|\le B$ for every $x\in E$ and every $N\ge m\ge1$.

Then $\sum a_n(x)b_n(x)$ converges uniformly on $E$. Its remainder satisfies

$$\sup_{x\in E}\left|\sum_{n=N+1}^{\infty}a_n(x)b_n(x)\right|\le B\sup_{x\in E}a_{N+1}(x).$$

Proof. Fix $x$ and apply the finite identity. Its monotonicity estimate gives $|\sum_{n=m}^N a_n(x)b_n(x)|\le Ba_m(x)$. Taking the supremum over $x$ gives a bound independent of the upper index. By assumption 2 this bound tends to zero as $m\to\infty$. Hence the partial sums are uniformly Cauchy. Completeness of the real numbers gives their pointwise limits, and the same uniform Cauchy bound gives uniform convergence: first choose a tail tolerance $\varepsilon/2$, then let the upper index tend to infinity for each $x$, obtaining an error at most $\varepsilon/2<\varepsilon$ for every $x$. Finally pass to the limit in the finite tail inequality to obtain the displayed remainder estimate. $\square$

Worked example 2: a sine series away from its troublesome endpoints

Let $0<\delta<\pi$ and $E_\delta=[\delta,2\pi-\delta]$. The addition identities imply

$$2\sin(x/2)\sin(nx)=\cos((n-1/2)x)-\cos((n+1/2)x).$$

Adding from $n=m$ to $N$ cancels all interior cosine terms. Therefore

$$\left|\sum_{n=m}^N\sin(nx)\right| \le\frac{2}{2|\sin(x/2)|}\le\frac1{\sin(\delta/2)}\qquad(x\in E_\delta).$$

The last inequality holds because $x/2\in[\delta/2,\pi-\delta/2]$, where sine is at least $\sin(\delta/2)>0$. With $a_n=1/n$, Theorem 2 proves that

$$\sum_{n=1}^{\infty}\frac{\sin(nx)}n$$

converges uniformly on $E_\delta$, with remainder after $N$ bounded by $1/[(N+1)\sin(\delta/2)]$. Each finite partial sum is continuous, so the sum is continuous on $E_\delta$. For each fixed $x\in(0,2\pi)$ some such closed interval contains $x$, giving pointwise convergence throughout that open interval.

Worked example 3: why the interval cannot be enlarged for free

The same sine series does not converge uniformly on $(0,2\pi)$, or on $[0,2\pi]$, even though the endpoint series are identically zero. For each positive integer $m$, choose $x_m=\pi/(6m)$. When $m\le n\le2m$, the angle $nx_m$ lies in $[\pi/6,\pi/3]$, so $\sin(nx_m)\ge1/2$. Also $n\le2m$, and therefore

$$\sum_{n=m}^{2m}\frac{\sin(nx_m)}n\ge\frac{m+1}{4m}>\frac14.$$

The uniform Cauchy criterion would make all sufficiently late blocks smaller than, for example, $1/8$ at every point of the domain. The displayed late blocks violate that requirement for every $m$. The parameter moves toward the endpoint as the tail index grows; checking only fixed parameters misses the failure.

Worked example 4: uniform convergence without a summable absolute majorant

On $[0,1]$, let $a_n(x)=x^n/n$ and $b_n=(-1)^{n-1}$. The weights are nonnegative and nonincreasing in $n$, including at $x=0$. Their supremum is $1/n\to0$, and alternating block sums have magnitude at most one. Hence

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}$$

converges uniformly on $[0,1]$, with tail at most $1/(N+1)$. But the supremum of the absolute value of its $n$th term is $1/n$. Any constants $M_n$ majorizing these terms on the whole interval must satisfy $M_n\ge1/n$, so $\sum M_n$ diverges. The M-test cannot apply here. Its failure is not a proof of failure of uniform convergence; a cancellation argument has supplied what absolute domination could not.

4. Abel's endpoint theorem: convergence at the endpoint first

Theorem 3. Suppose the numerical series $\sum_{n=0}^{\infty}c_n$ converges to $s$. Then the power series $\sum_{n=0}^{\infty}c_nr^n$ converges uniformly for $0\le r\le1$. If $F(r)$ denotes that sum, including $F(1)=s$, then $F$ is continuous on $[0,1]$ and

$$\lim_{r\uparrow1}\sum_{n=0}^{\infty}c_nr^n=\sum_{n=0}^{\infty}c_n.$$

Here $r^0$ is the constant polynomial one, also at $r=0$. The theorem assumes ordinary convergence at $r=1$; it does not manufacture it from an interior formula.

Proof by uniform tails. Given $\varepsilon>0$, numerical convergence supplies an index $m_0$ such that

$$\left|\sum_{n=m}^k c_n\right|<\varepsilon/2\qquad(k\ge m\ge m_0).$$

For each $r\in[0,1]$, use the finite summation identity with $a_n=r^n$ and $b_n=c_n$. These weights are nonnegative and nonincreasing, but their suprema do not tend to zero as $n\to\infty$; Theorem 2 alone would not apply. Instead it is the small coefficient-tail bound that matters. The identity gives

$$\begin{aligned} \left|\sum_{n=m}^N c_nr^n\right| &\le\frac\varepsilon2\left[r^N+\sum_{n=m}^{N-1}(r^n-r^{n+1})\right]\\ &=\frac\varepsilon2 r^m\le\frac\varepsilon2<\varepsilon. \end{aligned}$$

This is a uniform Cauchy estimate on the entire closed interval. Uniform completeness gives a uniform limit. Its partial sums are polynomials and therefore continuous; preservation of continuity under uniform limits gives continuity of $F$ on $[0,1]$. At $r=1$ its defining series is the assumed numerical sum $s$, so left continuity gives the claimed limit. $\square$

For a power series about $a$ with finite positive radius $R$, a convergent right endpoint is reduced to this theorem by $c_n=d_nR^n$ and $x-a=Rr$. A convergent left endpoint uses $c_n=d_n(-R)^n$ and $x-a=-Rr$. This scaling proves a one-sided approach from the interior; it makes no statement about points outside the radius.

5. Abel's weighted-average identity and a second endpoint proof

There is a useful complementary identity. Put $s_n=\sum_{k=0}^n c_k$. For $0\le r<1$,

$$F(r)=(1-r)\sum_{n=0}^{\infty}s_nr^n.$$

Justification. The finite identity is

$$\sum_{n=0}^N c_nr^n=(1-r)\sum_{n=0}^{N-1}s_nr^n+s_Nr^N.$$

It follows either by expanding the right side and subtracting adjacent coefficients, or by the finite summation formula. Because $s_n\to s$, there is a constant $K$ with $|s_n|\le K$ for all $n$. Thus $\sum |s_n|r^n\le K/(1-r)$ for $r<1$, and $|s_Nr^N|\le Kr^N\to0$. Passing to the limit in the finite identity is now justified.

The weights $(1-r)r^n$ are nonnegative and sum to one. Consequently

$$F(r)-s=(1-r)\sum_{n=0}^{\infty}(s_n-s)r^n.$$

Given $\varepsilon>0$, choose $N_0$ so that $|s_n-s|<\varepsilon/2$ for $n\ge N_0$. The tail of this expression has magnitude at most $\varepsilon/2$, since its weights sum to at most one. The initial finite part is at most

$$(1-r)\sum_{n=0}^{N_0-1}|s_n-s|,$$

which tends to zero as $r\uparrow1$. For $r$ sufficiently close to one it is below $\varepsilon/2$, proving $|F(r)-s|<\varepsilon$. This second proof makes the endpoint limit a weighted averaging of already convergent partial sums, not an unlicensed exchange of two limits.

Worked example 5: evaluating an endpoint after proving it exists

For each $0\le r<1$, the geometric series $\sum_{n=0}^{\infty}(-t)^n$ converges uniformly on $[0,r]$, since its terms are bounded by the summable constants $r^n$. Its sum is $1/(1+t)$. Termwise integration on that compact interval gives

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}r^n}{n}=\int_0^r\frac{dt}{1+t}=\log(1+r).$$

At $r=0$ the equality also holds, with both sides zero. Independently, the numerical alternating harmonic series converges by Dirichlet's test. Abel's endpoint theorem therefore applies and gives

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n=\lim_{r\uparrow1}\log(1+r)=\log2.$$

The proof has a necessary order: prove numerical endpoint convergence, prove the interior identity, and then identify the endpoint value through a justified limiting theorem.

6. Two boundaries that Abel's theorem does not remove

The converse is false. For $c_n=(-1)^n$, the interior power series is $1/(1+r)$ and has the finite limit $1/2$ as $r\uparrow1$. But $c_n$ does not tend to zero, so $\sum c_n$ diverges ordinarily. A finite radial limit can exist without convergence of the boundary series. Calling that radial limit an ordinary sum would change the meaning of convergence.

Endpoint differentiation is a different question. The series in Worked example 5 is uniformly convergent on $[0,1]$. For $r<1$, its derivative series is $\sum_{n=1}^{\infty}(-1)^{n-1}r^{n-1}=1/(1+r)$. At $r=1$ the derivative series has terms $(-1)^{n-1}$, which do not tend to zero. The function $\log(1+r)$ nevertheless has a finite left derivative $1/2$ at one. Thus uniform convergence of the function series at the endpoint neither guarantees convergence of the endpoint derivative series nor permits its termwise evaluation there. The termwise differentiation theorem inside the radius remains valid and has not been extended to its boundary.

7. Graded exercises

Foundation

P01. Verify the finite summation identity for $m=2,N=4$, with $a_n=1/n$ and $(b_2,b_3,b_4)=(3,-2,1)$. Compute the original sum and the two terms on the identity's right side exactly.

P02. Let $b_n$ repeat $3,-1,-2$. Prove that $\sum b_n/\sqrt n$ converges conditionally. Obtain an explicit remainder bound from the possible initial partial sums.

P03. Explain why the statement “bounded $b_n$ and nonincreasing $a_n\to0$ imply convergence of $\sum a_nb_n$” is false. Give a counterexample and identify the missing hypothesis.

Core

P04. In Dirichlet's test, is the condition $a_n\to0$ alone a replacement for nonnegative nonincreasing behavior? Take $b_n=(-1)^n$ and $a_n=[1+(-1)^n]/(2n)$, and analyze the proposed conclusion.

P05. For $E=[\pi/3,5\pi/3]$, give a uniform remainder bound for $\sum\sin(nx)/n$. Choose $N$ so the error after $N$ terms is strictly below $0.01$ on all of $E$.

P06. Prove that the sine series is not uniformly convergent on $[0,2\pi]$, even though each endpoint series converges. State one fixed positive tolerance that defeats the uniform Cauchy criterion.

P07. Define

$$G(r)=\sum_{n=0}^{\infty}\frac{(-1)^n r^n}{n+1}\qquad(0\le r\le1).$$

Prove uniform convergence, find $G(r)$ for $r>0$, specify $G(0)$, and evaluate $G(1)$. Give an error bound for the partial sum through index $N$ that is valid for every $r\in[0,1]$.

Synthesis and proof audit

P08. A student computes $\lim_{r\uparrow1}\sum_{n=0}^{\infty}(-1)^n r^n=1/2$ and concludes that $\sum_{n=0}^{\infty}(-1)^n=1/2$. Explain precisely which implication is unsupported and prove the ordinary series diverges.

P09. For $H(r)=\sum_{n=1}^{\infty}(-1)^{n-1}r^n/n$, compare (i) its left derivative at one, (ii) the limit of its derivatives as $r\uparrow1$ from inside the radius, and (iii) the series obtained by differentiating terms and then setting $r=1$. Justify all three answers and explain why no contradiction occurs.

P10. Suppose both $\sum_{n=0}^{\infty}c_n$ and $\sum_{n=0}^{\infty}(-1)^nc_n$ converge. Prove $\sum c_nx^n$ converges uniformly on the entire interval $[-1,1]$. Do not infer uniformity merely from pointwise convergence at both endpoints.

8. Full solutions

S01. The original sum is $3/2-2/3+1/4=13/12$. The block partial sums are $B_{2,2}=3$, $B_{2,3}=1$, and $B_{2,4}=2$. The final term in the identity is $a_4B_{2,4}=(1/4)2=1/2$. The difference sum is

$$(a_2-a_3)B_{2,2}+(a_3-a_4)B_{2,3} =(1/2-1/3)3+(1/3-1/4)1=1/2+1/12=7/12.$$

Adding the two contributions gives $1/2+7/12=13/12$, as required. No infinite limiting operation is involved.

S02. Initial partial sums of the periodic coefficients are $0,3,2$, repeated. Thus every block sum is a difference of two numbers in this set and has magnitude at most three. The weights $n^{-1/2}$ are positive, decrease, and tend to zero. Dirichlet's proof gives convergence and remainder after $N$ at most $3/\sqrt{N+1}$. Since $|b_n|\ge1$ at every index, the absolute series dominates $\sum n^{-1/2}$, which diverges by the $p$-series threshold. The original convergence is conditional.

S03. Take $b_n=1$ and $a_n=1/n$. The sequence $b_n$ is bounded and the weights decrease to zero, but the product series is harmonic and divergent. Its coefficient partial sums are $P_N=N$, which are unbounded. Dirichlet's hypothesis is a bound on these partial sums, not merely a bound on each term.

S04. The proposed weights are zero at odd indices and $1/n$ at even indices. They are nonnegative and tend to zero, but increase from zero to a positive number infinitely often, so they are not nonincreasing. Multiplication by $b_n=(-1)^n$ leaves the even terms positive and the odd terms zero. Hence

$$\sum_{n=1}^{\infty}a_nb_n=\sum_{k=1}^{\infty}\frac1{2k},$$

which diverges to $+\infty$. The coefficient partial sums are bounded, so the failure isolates the missing monotonicity of the weights.

S05. Here $\delta=\pi/3$ and $\sin(\delta/2)=\sin(\pi/6)=1/2$. The uniform tail bound is therefore $2/(N+1)$. For a strict error below $0.01=1/100$, it is enough that $N+1>200$. The choice $N=200$ gives $2/201<1/100$ at every $x$ in the stated interval. This is a uniform guarantee, not a claim about the exact largest error.

S06. Let $x_m=\pi/(6m)$ for $m\ge1$. Every $x_m$ belongs to $[0,2\pi]$. For $m\le n\le2m$, the angle lies in $[\pi/6,\pi/3]$, where sine is at least $1/2$, and $1/n\ge1/(2m)$. Thus the block sum is greater than $1/4$. If the series were uniformly convergent, its partial sums would be uniformly Cauchy with tolerance $1/8$. An eventual threshold would make every block starting beyond it smaller than $1/8$ at every point. Choosing $m$ beyond that threshold contradicts the displayed lower bound. The convergence of the two endpoint numerical series is irrelevant to this moving-point obstruction.

S07. For each $r\in[0,1]$, the magnitudes $r^n/(n+1)$ are nonnegative and nonincreasing in $n$. Their supremum over $r$ is $1/(n+1)\to0$, and alternating coefficient block sums have magnitude at most one. The uniform Dirichlet theorem gives uniform convergence. Its remainder after index $N$ is bounded by $1/(N+2)$ throughout $[0,1]$. For $r>0$, multiplying the series by $r$ produces the interior logarithm series,

$$rG(r)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}r^k}{k}=\log(1+r)\qquad(0<r<1).$$

Thus $G(r)=\log(1+r)/r$ there. At $r=0$ only the constant term remains, so $G(0)=1$; also $\log(1+r)/r\to1$ by the derivative of logarithm at one. Uniform convergence makes $G$ continuous at one, giving $G(1)=\log2$, and the same formula $\log(1+r)/r$ is valid also at $r=1$. No division by zero is used to define $G(0)$.

S08. For $0\le r<1$, the geometric formula gives the interior sum $1/(1+r)$ and hence radial limit $1/2$. Abel's theorem goes from an ordinarily convergent coefficient series to that radial limit. The student has used its converse, which was not proved and is false. The coefficient terms $(-1)^n$ do not approach zero. More explicitly, the partial sums through index $N$ are one for even $N$ and zero for odd $N$, so two subsequences have different limits. The ordinary series diverges.

S09. The previous worked example identifies $H(r)=\log(1+r)$ on $[0,1]$. Hence its left derivative at one is $1/2$. For each $0<r<1$, the interior power-series differentiation theorem gives $H'(r)=1/(1+r)$, whose limit as $r\uparrow1$ is again $1/2$. If terms are differentiated and $r=1$ is then substituted, the proposed series is $\sum_{n=1}^{\infty}(-1)^{n-1}$. Its terms do not approach zero and its partial sums alternate between one and zero, so it diverges. The two finite derivatives concern the function and an interior limit; the divergent expression concerns an endpoint interchange for which no derivative-series convergence hypothesis holds. The theorem on termwise differentiation was stated only inside the radius, so its valid conclusion is not contradicted.

S10. Apply Abel's uniform-tail proof to the convergent numerical series $\sum c_n$. For each $\varepsilon>0$, it supplies an index $N_+$ such that every tail $\sum_{n=m}^{N}c_nx^n$ has magnitude below $\varepsilon$ whenever $N\ge m\ge N_+$ and $x\in[0,1]$. Apply the same proof to $d_n=(-1)^nc_n$, obtaining a threshold $N_-$ valid for $r\in[0,1]$. If $x\in[-1,0]$, write $r=-x$, so $c_nx^n=d_nr^n$ and the second tail estimate applies. The common threshold $\max(N_+,N_-)$ therefore controls tails at every point of $[-1,1]$. The uniform Cauchy criterion proves uniform convergence on the whole interval. The proof combines two uniform half-interval estimates, rather than two endpoint values.

Summary and boundaries

Summation by parts transfers a weighted-sum problem to a bound on coefficient partial sums and the variation of the weights. Dirichlet cancellation can prove conditional or uniform convergence when the M-test cannot. Abel's endpoint theorem requires convergence of the numerical endpoint series and then proves uniform convergence toward it. Neither a finite radial limit nor uniform convergence of function values automatically justifies a boundary sum or boundary differentiation.