TheoremBase

Two auxiliary facts, splitting a finite sum at an index and transferring convergence from a shifted sequence back to the original, reduce each clause to the corresponding limit rules for the partial sums and to the Cauchy characterisation of convergence.

Proof

Each result cited is universally quantified over the data in its own statement. Two auxiliary facts come first.

(P1) Splitting. Let (dk)(d_{k}) be a sequence in R\mathbb{R} and m,q∈Nm,q\in\mathbb{N}. By Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §segment, [m+q]⊆N[m+q]\subseteq\mathbb{N}; so is the interval {m+1,…,m+q}\{m+1,\dots,m+q\} of Intervals of Natural Numbers §interval, whose elements kk satisfy 1≤m+1≤k1\le m+1\le k. By Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms §splitting, applied with mm, qq for its nn, mm to the restriction of (dk)(d_{k}) to [m+q][m+q] under the associative addition of R\mathbb{R}, and read as sums over intervals by Sums and Products over a Finite Set and over an Interval §intervals, ∑k=1m+qdk=∑k=1mdk+∑k=1qdm+k\sum_{k=1}^{m+q}d_{k}=\sum_{k=1}^{m}d_{k}+\sum_{k=1}^{q}d_{m+k}; and by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §interval-shift, applied with 11, qq, mm for its mm, nn, pp, ∑k=1qdm+k=∑k=m+1m+qdk\sum_{k=1}^{q}d_{m+k}=\sum_{k=m+1}^{m+q}d_{k}. Hence

∑k=1m+qdk=∑k=1mdk+∑k=m+1m+qdk,∑k=m+1m+qdk=∑k=1qdm+k.\sum_{k=1}^{m+q}d_{k}=\sum_{k=1}^{m}d_{k}+\sum_{k=m+1}^{m+q}d_{k},\qquad\sum_{k=m+1}^{m+q}d_{k}=\sum_{k=1}^{q}d_{m+k}.

(P2) Shifting back. Let (xn)(x_{n}) be a sequence in R\mathbb{R}, q∈Nq\in\mathbb{N} and L∈RL\in\mathbb{R} with (xq+n)n∈N(x_{q+n})_{n\in\mathbb{N}} converging to LL; then xn→Lx_{n}\to L. Indeed, let a real ε>0\varepsilon>0 be given; for this ε\varepsilon choose, by Convergent Sequences of Real Numbers §converges, N∈NN\in\mathbb{N} with ∣xq+n−L∣<ε|x_{q+n}-L|<\varepsilon for every n≥Nn\ge N. Let n∈Nn\in\mathbb{N} with n≥q+Nn\ge q+N. As q<q+N≤nq<q+N\le n, n=q+dn=q+d for some d∈Nd\in\mathbb{N} by Arithmetic and Order of the Natural Numbers §difference, and q+N≤q+dq+N\le q+d gives N≤dN\le d by Arithmetic and Order of the Natural Numbers §order. Hence ∣xn−L∣=∣xq+d−L∣<ε|x_{n}-L|=|x_{q+d}-L|<\varepsilon. Since ε\varepsilon was arbitrary and q+N∈Nq+N\in\mathbb{N}, xn→Lx_{n}\to L by Convergent Sequences of Real Numbers §converges.

Clause linear. Let A=∑k=1∞akA=\sum_{k=1}^{\infty}a_{k} and B=∑k=1∞bkB=\sum_{k=1}^{\infty}b_{k}; then sn→As_{n}\to A and tn→Bt_{n}\to B by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit. For n∈Nn\in\mathbb{N}, by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §partial-sums the nn-th partial sum of (ak+bk)(a_{k}+b_{k}) is ∑k=1n(ak+bk)=sn+tn\sum_{k=1}^{n}(a_{k}+b_{k})=s_{n}+t_{n}, by Sums and Products over a Finite Set and over an Interval §intervals and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §termwise, and that of (λak)(\lambda a_{k}) is ∑k=1nλak=λsn\sum_{k=1}^{n}\lambda a_{k}=\lambda s_{n}, by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §distributive. By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic, sn+tn→A+Bs_{n}+t_{n}\to A+B and λsn→λA\lambda s_{n}\to\lambda A. Hence both series converge by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges, with sums A+BA+B and λA\lambda A by The Limit of a Convergent Sequence §limit.

Clause terms-null. Let S=∑k=1∞akS=\sum_{k=1}^{\infty}a_{k}, so sn→Ss_{n}\to S by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit, and (sn+1)n∈N(s_{n+1})_{n\in\mathbb{N}} converges to SS by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §tails. For n∈Nn\in\mathbb{N}, sn+1=sn+an+1s_{n+1}=s_{n}+a_{n+1} by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §partial-sums and Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §interval-recursion, so a1+n=an+1=sn+1−sna_{1+n}=a_{n+1}=s_{n+1}-s_{n}. By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic, sn+1−sn→S−S=0s_{n+1}-s_{n}\to S-S=0. Thus (a1+n)n∈N(a_{1+n})_{n\in\mathbb{N}} converges to 00, and by (P2) with q=1q=1, (ak)(a_{k}) is a null sequence in the sense of Convergent Sequences of Real Numbers §converges.

Clause cauchy. For m,n∈Nm,n\in\mathbb{N} with m<nm<n there is q∈Nq\in\mathbb{N} with n=m+qn=m+q by Arithmetic and Order of the Natural Numbers §difference, so (P1) with (ak)(a_{k}) for (dk)(d_{k}) and Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §partial-sums give sn=sm+∑k=m+1naks_{n}=s_{m}+\sum_{k=m+1}^{n}a_{k}, that is, (∗)(\ast) ∑k=m+1nak=sn−sm\sum_{k=m+1}^{n}a_{k}=s_{n}-s_{m}. By Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences §cauchy, the series converges if and only if (sn)(s_{n}) is a Cauchy sequence as in Cauchy Sequences of Real Numbers §cauchy. Suppose it is, and let a real ε>0\varepsilon>0 be given; for this ε\varepsilon choose NN by Cauchy Sequences of Real Numbers §cauchy. For m,n∈Nm,n\in\mathbb{N} with N≤m<nN\le m<n, also N≤nN\le n, so ∣∑k=m+1nak∣=∣sn−sm∣<ε\big|\sum_{k=m+1}^{n}a_{k}\big|=|s_{n}-s_{m}|<\varepsilon by (∗)(\ast). Conversely, suppose the condition of the clause, and let a real ε>0\varepsilon>0 be given; for this ε\varepsilon choose NN by the condition. Let m,n∈Nm,n\in\mathbb{N} with m,n≥Nm,n\ge N. By Arithmetic and Order of the Natural Numbers §trichotomy, m=nm=n, in which case ∣sm−sn∣=0<ε|s_{m}-s_{n}|=0<\varepsilon; or m<nm<n, in which case ∣sm−sn∣=∣sn−sm∣=∣∑k=m+1nak∣<ε|s_{m}-s_{n}|=|s_{n}-s_{m}|=\big|\sum_{k=m+1}^{n}a_{k}\big|<\varepsilon by (∗)(\ast); or n<mn<m, in which case ∣sm−sn∣=∣∑k=n+1mak∣<ε|s_{m}-s_{n}|=\big|\sum_{k=n+1}^{m}a_{k}\big|<\varepsilon by (∗)(\ast) with mm and nn exchanged. So (sn)(s_{n}) is a Cauchy sequence by Cauchy Sequences of Real Numbers §cauchy. Thus the condition of the clause holds if and only if (sn)(s_{n}) is a Cauchy sequence, and, by what was shown first, this holds if and only if the series converges; these are the two equivalences of the clause.

Clause shift. For n∈Nn\in\mathbb{N} let un=∑k=1nap+ku_{n}=\sum_{k=1}^{n}a_{p+k}, the nn-th partial sum of (ap+k)k∈N(a_{p+k})_{k\in\mathbb{N}} by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §partial-sums. By (P1) with (ak)(a_{k}), pp and nn for (dk)(d_{k}), mm and qq, sp+n=sp+uns_{p+n}=s_{p}+u_{n} for every n∈Nn\in\mathbb{N}. Suppose ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges, with sum SS; then sn→Ss_{n}\to S by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit, so (sn+p)n∈N(s_{n+p})_{n\in\mathbb{N}}, which is (sp+n)n∈N(s_{p+n})_{n\in\mathbb{N}}, converges to SS by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §tails. The constant sequence (sp)n∈N(s_{p})_{n\in\mathbb{N}} converges to sps_{p} by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant, so un=sp+n−sp→S−spu_{n}=s_{p+n}-s_{p}\to S-s_{p} by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic. Hence ∑k=1∞ap+k\sum_{k=1}^{\infty}a_{p+k} converges by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges, with sum S−spS-s_{p} by The Limit of a Convergent Sequence §limit, which is the stated equation. Conversely, suppose ∑k=1∞ap+k\sum_{k=1}^{\infty}a_{p+k} converges, with sum UU; then un→Uu_{n}\to U by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit, so sp+n=sp+un→sp+Us_{p+n}=s_{p}+u_{n}\to s_{p}+U by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic, and sn→sp+Us_{n}\to s_{p}+U by (P2) with q=pq=p. Hence ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges, and the equation holds by the first part.

Clause tails. Let S=∑k=1∞akS=\sum_{k=1}^{\infty}a_{k}, so sq→Ss_{q}\to S by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit. For q∈Nq\in\mathbb{N}, clause shift above with qq for pp shows that ∑k=1∞aq+k\sum_{k=1}^{\infty}a_{q+k} converges, with ∑k=1∞aq+k=S−sq\sum_{k=1}^{\infty}a_{q+k}=S-s_{q}. The constant sequence (S)q∈N(S)_{q\in\mathbb{N}} converges to SS by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant, so S−sq→S−S=0S-s_{q}\to S-S=0 by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic; that is, the sequence of tails is a null sequence by Convergent Sequences of Real Numbers §converges.

Clause order. Let A=∑k=1∞akA=\sum_{k=1}^{\infty}a_{k} and B=∑k=1∞bkB=\sum_{k=1}^{\infty}b_{k}; then sn→As_{n}\to A and tn→Bt_{n}\to B by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit. For n∈Nn\in\mathbb{N}, sn≤tns_{n}\le t_{n} by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §partial-sums and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §comparison, applied in the ordered field R\mathbb{R} to the restrictions of (ak)(a_{k}) and (bk)(b_{k}) to [n][n]. Hence A≤BA\le B by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §order with K=1K=1.

Clause telescoping. For n∈Nn\in\mathbb{N}, sn=∑k=1n(ck+1−ck)=cn+1−c1s_{n}=\sum_{k=1}^{n}(c_{k+1}-c_{k})=c_{n+1}-c_{1} by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §partial-sums and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality §telescoping, applied to the restriction of (ck)(c_{k}) to [n+1][n+1]. Suppose (ck)(c_{k}) is convergent, with limit C=lim⁡k→∞ckC=\lim_{k\to\infty}c_{k} as in The Limit of a Convergent Sequence §limit; then (cn+1)n∈N(c_{n+1})_{n\in\mathbb{N}} converges to CC by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §tails, so sn=cn+1−c1→C−c1s_{n}=c_{n+1}-c_{1}\to C-c_{1} by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic. Hence ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges, with sum C−c1C-c_{1} by The Limit of a Convergent Sequence §limit. Conversely, suppose ∑k=1∞ak\sum_{k=1}^{\infty}a_{k} converges, with sum SS; then sn→Ss_{n}\to S by Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence §converges and The Limit of a Convergent Sequence §limit, so c1+n=cn+1=sn+c1→S+c1c_{1+n}=c_{n+1}=s_{n}+c_{1}\to S+c_{1} by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic. By (P2) with q=1q=1, cn→S+c1c_{n}\to S+c_{1}, so (ck)(c_{k}) is convergent by Convergent Sequences of Real Numbers §converges, and the formula for the sum holds by what was shown first.

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