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Tauber's theorem: an Abel-summable series with $\iota(n+1)a_n\to0$ converges ordinarily to its Abel sum

Statement

Let $\sum_{n\ge0}a_n$ be Abel summable to $s$ . If

$\iota(n+1)a_n\longrightarrow0,$

then its ordinary partial sums converge to $s$ .

Facts & Assumptions

Given: The Abel sum $A(x):=\sum_{n\ge0}a_nx^n\to s$ as $x\uparrow1$ and the stated Tauber condition.

[L1]

The block lemma supplies uniform bounds for the weighted middle and tail when $x_N:=1-1/\iota(N+1)$ (If $\iota(n+1)a_n\to0$ , short multiplicative blocks of the coefficients have uniformly small sums).

[L2]

The Archimedean reciprocal property gives a reciprocal below every positive tolerance. Canonical naturals increase and reciprocation reverses positive order, so every later reciprocal remains below that tolerance; hence $1/\iota(N+1)\to0$ and $x_N\uparrow1$ . Abel summability then gives $A(x_N)\to s$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Abel summability by $\lim_{x\uparrow1}\sum a_nx^n$ and Cesaro summability by the Cesaro means of the partial sums, Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

Write $S_N:=\sum_{n=0}^{N}a_n$ . For $N\ge1$ , one has $S_N-A(x_N)=\sum_{n=0}^{N}a_n(1-x_N^n)-\sum_{n>N}a_nx_N^n$ .

givenalgebra

2.1

Given $\varepsilon>0$ , choose $N_0$ from [L1]. The part of the first sum with $n<N_0$ tends to $0$ because it is finite and $x_N\to1$ ; the remaining part and the tail have absolute value at most $\varepsilon$ each by [L1].

step 1.1L1choose

3.1

Hence $S_N-A(x_N)\to0$ . Since $A(x_N)\to s$ by [L2], it follows that $S_N\to s$ .

step 2.1L2∎

Depends on

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Sources

Tauberian theorem, Encyclopedia of Mathematics (standard reference, not scraped)
Tauberian theorems, Encyclopedia of Mathematics (standard reference, not scraped)

Tauber's theorem: an Abel-summable series with ι(n+1)an→0\iota(n+1)a_n\to0ι(n+1)an​→0 converges ordinarily to its Abel sum