Tauberian theorems for general power series methods

by

R.Kiesel and U.Stadtmüller

1 Introduction and main result

Let us assume throughout that (p_{n}) denotes a sequence of reals which satisfies

(1.1) \{  \begin{matrix}p_{0} > 0, p_{n} ≥ 0, n = 1, 2, …, such that the power series \\ p(t) = ∑_{n=0}^{∞} p_{n}t^{n} has radius of convergence R ∈ (0, ∞]. \end{matrix}

For real sequences (s_{n}) with increments a_{n} = s_{n} − s_{n−1}, n ≥ 0, (s_{−1} = 0) , we consider the power-series method of summability (P) , where we say

(1.2)  \{  \begin{matrix}s_{n} → s(P) if p_{s}(t) = ∑_{n=0}^{∞} s_{n}p_{n}t^{n} converges for |t| < R \\ and if σ_{p}(t) = \frac{p_{s}(t)}{p(t)} → s as t → R − . \end{matrix}

The power series methods (P) contain the so–called (J_{p}) –methods ( R = 1 ) and the Borel–type methods (B_{p}) ( R = ∞). We consider only regular (P) –methods, i.e. s_{n} → s implies s_{n} → s(P) . By Theorem 5 in [5], p.49, we have regularity iff

(1.3) \{ \begin{matrix}∑_{k=0}^{n} p_{k}R^{k} → ∞, (n → ∞) in case R < ∞, \\ ( in particular P_{n} = ∑_{k=0}^{n} p_{k} → ∞, (n → ∞) if R = 1,) \\ p(t) is not a polynomial, i.e. p_{n} ≠ 0 infinitely often in case R = ∞. \end{matrix}

Here we are interested in the converse conclusion, namely s_{n} → s(P) implies s_{n} → s, which can only be valid if further conditions, so called Tauberian conditions are satisfied by (s_{n}) . These so-called Tauberian–theorems for power series methods have a long history, see e.g the books [5,14,23], and found new attention recently in the papers [6,18,19,20] and [8,9,10,11,12]. The latter papers contain certain o –Tauberian theorems for all power series methods in question and O –Tauberian theorems, if the weight sequence (p_{n}) can be interpolated by a logarithmico–exponential function g(.) , see e.g. [4], i.e.

(1.4)  p_{n} ∼ exp{g(n)}.

In order to state these results we need the following quantities:

(1.5)  Δ_{n} = inf_{0<t<R} p(t)t^{−n}, n = 0, 1, 2, ... ,

which by the results in [2,12] can be written as

(1.6) Δ_{n} = p(t_{n})t_{n}^{−n} for n = 0, 1, 2, … with some sequence 0 < (t_{n}) ↗ R.

Then an o –Tauberian condition, valid for all methods satisfying (1.1) and (1.3) is given, see [9,10,12], by

(1.7)  lim_{n→∞} max_{n≤m≤n+l(n)} |s_{m+1} − s_{n}| = 0

where (l(n)) is a sequence of integers s.t.

(1.8)  ∑_{k=n}^{n+l(n)} \frac{Δ_{k}}{p_{k}} ≥ 1, n ∈ N I .

Under the additional asumption (1.4) we can weaken this Tauberian condition to the following O –condition, see [11,12]:

(1.9)  ω(λ) := lim sup_{n→∞} max_{n≤m≤n+l(n,λ)} |s_{m+1} − s_{n}| → 0, (λ → 1+),

where (l(n, λ)) now is for each λ > 1 a sequence of integers s.t.

(1.8′)  ∑_{k=n}^{n+l(n,λ)} \frac{Δ_{k}}{p_{k}} ≥ λ − 1, n ∈ N I .

If we denote in (1.8) or (1.8′) the shortest possible by sequences (l^{∗}(n)), (l^{∗}(n, λ)) resp. (for a discussion see [10,11]), then we can equivalently choose the sequences (cl^{∗}(n)), (cl^{∗}(n, λ)) resp. for any c > 0 , provided (1.4) holds. Hence in case of:

(i) the Abel method with p_{n} = 1, R = 1 : Δ_{n} ∼ en and l(n, λ) can be choosen as (λ − 1)n, see e.g. [5,16].

More general, if p_{n} = R(n) with a regularly varying function R(.), see [1,17], with index α > −1 , then again Δ_{n} ∼ c(α)n and l(n, λ) = (λ − 1)n is a possible choice, see e.g. [6,10,11].

(ii) the Borel method with p_{n} = \frac{1}{n!}, R = ∞:

Δ_{n} = (\frac{e}{n})^{n} and l(n, λ) can be choosen as (λ − 1)\sqrt{n}, see e.g. [5].

In all this special cases a one–sided version of (1.9) is a sufficient Tauberian condition (TC) as well, see [6,16,18,21,22], i.e.

(1.9′)  ω(λ) := lim inf_{n→∞} min_{n≤m≤n+l(n,λ)}(s_{m+1} − s_{n}) satisfies lim_{λ→1+} ω(λ) ≥ 0 ,

where (l(n, λ)) is defined as in (1.8′) .

The proofs of all one–sided results contain the following steps:

(1) the Vijayaraghavan–type conclusion: s_{n} → s(P) plus (TC) implies s_{n} = O (1) ;

(2) the weaker summability method step: s_{n} → s(P) plus s_{n} = O (1) implies s_{n} → s(S) , where (S) stands for a (or a class of) weaker summability–method(s);

(3) the direct argument:

s_{n} → s(S) plus (TC) implies s_{n} → s.

The main result in this paper is a Vijayaraghavan–type theorem giving the conclusion (1) for all power series methods in question. The proof uses the general properties of the sequences (Δ_{n}) and (t_{n}) and avoids in contrast to the literature special asymptotic considerations.

THEOREM 1 Let be given a regular power–series method (P) , obeying (1.1) and (1.3) with a sequence of weights (p_{n}) . Suppose further, that (s_{n}) is a sequence of reals, satisfying

s_{n} = O (1) (P), i.e. \frac{p_{s}(t)}{p(t)} = O (1) , (t → R−)

and

(1.9′′)  ω(λ) = lim inf_{n→∞} min_{n≤m≤n+l_{n}}(s_{m+1} − s_{n}) > −∞,

where (l_{n}) is a sequence of nonnegative integers with

(1.8′′)  lim inf_{n→∞} ( ∑_{k=n}^{n+l_{n}} \frac{p_{k}}{Δ_{k}} ) > 0.

Then we have s_{n} = O (1) .

Remark 1:

(i) In (1.9′′) we may use any of the sequences (l(n)), (l(n, λ)), (l^{∗}(n)), (l^{∗}(n, λ)) for any λ > 1 .

(ii) The following cases have been discussed in the literature: (In what follows c > 0 stands for some constant.)

(a) the Logarithmic method, [13]:

p_{n} = \frac{1}{n}, R = 1, l(n) = c n log n.

(b) the Abel method, [16,21], or more generally (J_{p}) –methods with polynomial weights, [6,18]: p_{n} = 1 , or p_{n} = R(n) , R(.) some regularly varying function with Index α > −1 . Then we have R = 1 and l(n) = cn.

(c) (J_{p}) –methods with exponential increasing weights, [7]: p_{n} = e^{n^{α}}, 0 < α < 1, R = 1, l(n) = cn^{1− \frac{α}{2}} .

(d) the Borel method, [22]:

p_{n} = \frac{1}{n!}, R = ∞, l(n) = c\sqrt{n}.

(e) the Generalized Borel–methods, [3]:

p_{n} = \frac{α}{Γ(αn + β)} (with t replaced by t^{α}), α, β > 0, l(n) = c\sqrt{n}.

(f) (B_{p}) –methods, [7]:

p_{n} = e^{−n^{β}}, 1 < β < 2, R = ∞, l(n) = cn^{1− \frac{β}{2}} .

In section 2 we shall prove Theorem 1, whereas in section 3 we give a Tauberian theorem under the one–sided condition (1.9′) for all power series methods, provided (1.4) is valid.

2 Proof of Theorem 1

The main tools for the proof are a Vijayaraghavan type theorem and the results about the quantities Δ_{n} (see (1.5)) stated in [2,9,12]. In what follows we consider the method of summation ( χ) defined by

(2.1)  χ(t) = ∑_{n=0}^{∞} c_{n}(t)s_{n} → s

for t → ∞ and (s_{n}) a sequence of reals. We suppose that

(2.2)  c_{n}(t) ≥ 0 (t > 0, n ≥ 0), c_{n}(t) → 0, ∑_{n=0}^{∞} c_{n}(t) → 1, (t → ∞)

so that (χ) is regular. Then we have the following Vijayaraghavan type theorem.

LEMMA 1 Consider the summability method (2.1) satisfying (2.2) and a real sequence (s_{n}) . If there exists a positive, strictly increasing function Φ(.) on (0, ∞) s.t

(2.3)  Φ(t + 1) − Φ(t) ≤ 2 and Φ(t) → ∞, (t → ∞)

and

(2.4) ∑_{n=0}^{M} c_{n}(t) → 0 , as Φ(t) − Φ(M) → ∞, t > M → ∞;

(2.5) ∑_{n=N}^{∞} c_{n}(t)(Φ(n) − Φ(N)) → 0 , as Φ(N) − Φ(t) → ∞, N > t → ∞;

(2.6) and if there exist positive constants a, b s.t. for u > t > 0

s(u) − s(t) > −a{Φ(u) − Φ(t)} − b,

where s(t) = s_{n} for n ≤ t < n + 1,

then s_{n} = O (1) (χ) , implies s_{n} = O (1) .

Proof: One has to modify slightly the proof of Theorem 238 in [5] or of Lemma 1 in [15], but we demand only the boundedness of Φ(t + 1) − Φ(t) rather then the convergence of this difference to zero.

Proof of Theorem 1: We want to use Lemma 1, therefore it is useful to rescale the (P) –methods. Starting with the sequences (t_{n}) as in (1.6), we define a nondecreasing function h : [0, ∞) → R I s.t. h(n) = log t_{n}, n ∈ N I , hence lim_{t→∞} h(t) = log R. Then we

put for a given power series method (P)

c_{n}(t) = \frac{p_{n} exp\{nh(t)\}}{p(e^{h(t)})} , t > 0, n = 0, 1, 2, …

By (1.1) and (1.3) c_{n}(t) satisfies (2.2). Next we define a function Φ(.) , beginning

with f(t) = ∑_{k=0}^{[t]} \frac{p_{k}}{Δ_{k}} , where as usual [t] = n, if n ≤ t < n + 1 for some n ∈ N_{0} I .

Then we replace the step function f(.) by a strictly increasing function Φ : (0, ∞) → (0, ∞) , which interpolates f(.) at all points t = n, where p_{n} > 0 . Φ(.) satisfies (2.3) as desired, since

\frac{p_{k}}{Δ_{k}} ≤ 1

and

∑_{k=0}^{∞} \frac{p_{k}}{Δ_{k}} = ∞

by Lemma 1 in [9,12]. With these quantities we check the conditions (2.4) – (2.6).

ad (2.4) :

Observe that by (1.6), see also [9,12]

p (e^{h(n)}) e^{−nh(n)} = Δ_{n} ≤ Δ_{k}e^{(k−n)h(k)},

i.e. Δ_{k} ≥ Δ_{n}e^{−(k−n)h(k)} for all k, n ∈ N I . Hence for t > M > 0, M ∈ N I , we obtain by the monotonicity of h(.) :

\begin{aligned}Φ(t) − Φ(M) & ≤ & ∑_{n=M+1}^{[t]+1} \frac{p_{n}}{Δ_{n}} ≤ \frac{1}{Δ_{M}} ∑_{n=M+1}^{[t]} p_{n}e^{(n−M)h(n)} + 1 \\ & ≤ & \frac{1}{Δ_{M}} ∑_{n=M+1}^{[t]} p_{n}e^{(n−M)h(t)} + 1 = \frac{e^{−Mh(t)}}{Δ_{M}} ∑_{n=M+1}^{[t]} p_{n}e^{nh(t)} + 1. \end{aligned}

So from (2.4) we can assume that the first term tends to infinity as t > M → ∞. This implies (use again the monotonicity of h(.) and (1.6)):

∑_{n=0}^{M} c_{n}(t) = ∑_{n=0}^{M} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} = ∑_{n=0}^{M} \frac{p_{n}e^{nh(M)+n(h(t)−h(M))}}{p (e^{h(t)})}

≤ \frac{e^{M(h(t)−h(M))}}{p (e^{h(t)})} ∑_{n=0}^{M} p_{n}e^{nh(M)}

≤ \frac{e^{−Mh(M)} ∑_{n=0}^{∞} p_{n}e^{nh(M)}}{p (e^{h(t)}) e^{−Mh(t)}}

≤ \frac{Δ_{M}}{e^{−Mh(t)} ∑_{n=M+1}^{[t]} p_{n}e^{nh(t)}}

≤ \frac{1}{Φ(t) − Φ(M) − 1} → 0, (M → ∞),

i.e (2.4) is valid.

ad (2.5) :

Consider the sequence l(n) , satisfying

∑_{k=n}^{n+l(n)−1}\frac{p_{k}}{Δ_{k}} < 1 ≤ ∑_{k=n}^{n+l(n)}\frac{p_{k}}{Δ_{k}} < 2.

and put n_{0} = [t] − 1, n_{ν+1} = n_{ν} + l(n_{ν}) ; ν = 0, 1, 2, … and α = α(t, N) ∈ N I , s.t. n_{α+1} > N ≥ n_{α} for given t > N ∈ N I . Then we obtain

Φ(N) − Φ(t) ≤ ∑_{n=[t]+1}^{N} \frac{p_{n}}{Δ_{n}} ≤ ∑_{ν=0}^{α} ∑_{n=n_{ν}+1}^{n_{ν+1}} \frac{p_{n}}{Δ_{n}} ≤ 2(α + 1).

In (2.5) we assume that Φ(N) − Φ(t) → ∞, (t → ∞) , hence α → ∞, (t → ∞) . Moreover using the definition and the monotonicity of h(.) and (1.6), we obtain for [t] − 1 ≤ s:

\begin{aligned}∑_{n=s+1}^{∞} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} ∑_{k=s+1}^{n} \frac{p_{k}}{Δ_{k}} & = ∑_{k=s+1}^{∞} \frac{p_{k}}{Δ_{k}} ∑_{n=k}^{∞} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} \\ & = \frac{1}{p (e^{h(t)})} ∑_{k=s+1}^{∞} \frac{p_{k}}{Δ_{k}} (\frac{e^{h(t)}}{e^{h(k)}} )^{k} ∑_{n=k}^{∞} p_{n}e^{nh(k)} (\frac{e^{h(k)}}{e^{h(t)}} )^{k−n} \end{aligned}

\begin{matrix}≤ \frac{1}{p (e^{h(t)})} ∑_{k=s+1}^{∞} p_{k}e^{kh(t)} \frac{p (e^{h(k)}) e^{−kh(k)}}{Δ_{k}} \\ = ∑_{k=s+1}^{∞} \frac{p_{k}e^{kh(t)}}{p (e^{h(t)})} . \end{matrix}

Choosing especially s = [t] − 1 and n_{ν} as above, we find by the choice of the n_{ν} :

1 ≥ ∑_{n=[t]+1}^{∞} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} ∑_{k=[t]+1}^{n} \frac{p_{k}}{Δ_{k}} = ∑_{ν=0}^{∞} ∑_{n=n_{ν}+1}^{n_{ν+1}} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} ∑_{k=n_{0}+1}^{n} \frac{p_{k}}{Δ_{k}}

≥ ∑_{ν=0}^{∞} ∑_{n=n_{ν}+1}^{n_{ν+1}} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} ∑_{k=n_{0}+1}^{n_{ν}} \frac{p_{k}}{Δ_{k}}

≥ ∑_{ν=0}^{∞} ν ∑_{n=n_{ν}+1}^{n_{ν+1}} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} .

Using the last two chains of inequalities, we conclude for s = N :

∑_{n=N}^{∞} c_{n}(t)(Φ(n) − Φ(N)) = ∑_{n=N+1}^{∞} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} ∑_{k=N+1}^{n} \frac{p_{k}}{Δ_{k}}

≤ ∑_{n=N+1}^{∞} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})} ≤ ∑_{ν=α}^{∞} ∑_{n=n_{ν}+1}^{n_{ν+1}} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})}

≤ ∑_{ν=α}^{∞} \frac{ν}{α} ∑_{n=n_{ν}+1}^{n_{ν+1}} \frac{p_{n}e^{nh(t)}}{p (e^{h(t)})}

≤ \frac{1}{α} → 0, (t → ∞).

ad (2.6):

By assumption (1.9′′) there exists some c > 0 s.t.

s_{m+1} − s_{n} > −c for all m ∈ [n, n + l_{n}],

and n large enough. With s(t) = s_{n}, l(t) = l_{n} if n ≤ t < n + 1, n ∈ N_{0} I , we obtain

s(u) − s(t) > −c for t ≤ u ≤ [t] + l(t),

hence

s(u) − s(t) > −2c for t ≤ u ≤ t + l(t).

For given u > t we define t_{0} = t, t_{ν+1} = t_{ν} + l(t_{ν}) with ν = 0, 1, …, r, and r ∈ N I s.t. t_{r+1} > u ≥ t_{r} and obtain

s(u) − s(t) = ∑_{ν=0}^{r−1}{s(t_{ν+1}) − s(t_{ν})} + s(u) − s(t_{r}) ≥ −(r + 1)2c.

By the definition of the sequence l_{n}, see (1.8′′) , there exists some constant δ > 0 , s.t.

0 < δ ≤ ∑_{k=n}^{n+l_{n}} \frac{p_{k}}{Δ_{k}}

for all large n and therefore

\begin{aligned}r & ≤ \frac{1}{δ} ∑_{ν=0}^{r−1} ∑_{k=[t_{ν}]+1}^{[t_{ν+1}]} \frac{p_{k}}{Δ_{k}} = \frac{1}{δ} ∑_{ν=0}^{r−1}\{Φ([t_{ν+1}]) − Φ([t_{ν}])\} \\ & = \frac{1}{δ}\{Φ([t_{r}]) − Φ([t_{0}])\} ≤ \frac{1}{δ}\{Φ(u) − Φ(t) + 2\}, \end{aligned}

by (2.3) and the monotonicity of Φ(.) . Altogether we obtain for u > t (large enough)

s(u) − s(t) ≥ −2c(r + 1) ≥ −2\frac{c}{δ} \{Φ(u) − Φ(t) + 2\} − 2c,

i.e. a = \frac{2c}{δ}, b = \frac{4c}{δ} + 2c.

Hence Lemma 1 implies Theorem 1.

3 A one–sided Tauberian Theorem

Sharpening the Tauberian–condition of Theorem 1 we get a one–sided Tauberian theorem.

THEOREM 2. Given a regular power–series method (P) with a sequence of weights (p_{n}) satisfying (1.1),(1.3) and (1.4). Suppose further that (s_{n}) is a sequence of reals satisfying

s_{n} → s(P) and the Tauberian–condition (1.9′),

then we have s_{n} → s.

We only sketch the Proof: By Theorem 1 we find that s_{n} = O (1) . The rest of the proof follows by techniques used elsewhere repeatedly so we only outline the main points. W.l.o.g we assume R = 1 or R = ∞.

(a) In case R = 1 and \frac{log P_{n}}{log n} = O (1) , s_{n} = O (1) is a Tauberian condition for

s_{n} → s(P) to s_{n} → s(M_{p}) , i.e.

\frac{∑_{k=0}^{n} s_{k}p_{k}}{∑_{k=0}^{n} p_{k}} → s

(see [10,20], observe that P_{n} is regularly varying under (1.4)). Next s_{n} → s(M_{p}) and (1.9′) imply s_{n} → s, see [6] Theorem 6.1 or [18] Theorem 3.1.

(b) In the other cases, we can choose l(n, λ) = (λ − 1)/\sqrt{|g′′(n)|}. We are now in the situation discussed in [11] R = 1 and [12] R = ∞ and we conclude as therein that s_{n} → s(P) and s_{n} = O (1) imply that s_{n} → s(P_{α}) for all α > −1 , where (P_{α}) is the power series method with weights p_{n}(α) = p_{n} exp{αg(n)} in case R = 1 and p_{n}(α) = p_{n} exp{−αg(n)} in case R = ∞ respectively. Using the

asymptotic calculations in [11] and [12] we obtain with s(t) = \{ \begin{matrix}s_{n} & t = n ∈ N_{0} I \\ 0 & t < 0 \end{matrix}

and linear elsewhere, that

(3.2) \sqrt{\frac{|g′′(t)|(α + 1)}{2π}} ∫_{−∞}^{∞} exp\{−(α + 1)|g′′(t)|x^{2}\}s(x + t) dx → s, (t → ∞).

We may now put w.l.o.g. s = 0 and assume that s_{n} ↛ 0 , i.e. there exists some sequence n_{k} s.t. s_{n_{k}} → δ > 0 . By (1.9′) we find that for λ − 1 > 0 , but small enough, k large

s(n_{k} + x) ≥ \frac{δ}{2} for |x| ≤ (λ − 1)/\sqrt{|g′′(n_{k})|}.

But since the integral

\begin{matrix}\sqrt{\frac{|g′′(t)|(α + 1)}{2π}} ∫|x| ≤ \frac{(λ − 1)\sqrt{α + 1}}{\sqrt{(α + 1)|g′′(t)|}} exp\{−(α + 1)|g′′(t)|x^{2}\} dx \\ = \frac{1}{\sqrt{2π}} ∫|x| ≤ (λ − 1)\sqrt{α + 1} exp\{−\frac{x^{2}}{2} \} dx → 1, (α → ∞) \end{matrix}

uniformly in t, we obtain a contradiction to (3.2) and hence Theorem 2.

Remark 2:

(i) In case \frac{log P_{n}}{log n} = O (1) one can generalize the result to a somewhat larger

class {p_{n}} than considered, see [18] for details.

(ii) With our Theorem 2 we obtain all one–sided oscillation Tauberian theorems known in the literature for the special cases we gave in Remark 1.

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