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UDK 517.521

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SEGMENTS OF EXPONENTIAL SERIES AND REGULARLY VARYING SEQUENCES

Slavko Simić

Abstract. The task of this paper is to investigate asymptotic behavior of segments of exponential series defined as

T_{λ}(x) := ∑_{n<λx} \frac{c_{n}}{n!} x^{n}, λ ∈ R^{+}, x → ∞,

where (c_{n})_{n∈N} belongs to the set of regularly varying sequences in Karamata sense of arbitrary index. Precise results are obtained.

Introduction

Karamata’s class R_{α} of regularly varying functions with index α ∈ R consists of all functions a(x) representable in the form a(x) = x^{α}l(x), where l(x) is from the class of so-called slowly varying functions, i.e. defined on positive part of real axis, positive, measurable and satisfying lim_{x→∞} \frac{l(sx)}{l(x)} = 1, for each s > 0.

According to [3], we could treat regularly varying sequnece (c_{n}) of index α as generated from some a ∈ R_{α}, i.e. c_{n} = n^{α}l(n), n ∈ N.

After seventy years, Karamat’s theory is very well developed and found applications in different parts of analysis. An excellent survey of results could be found in [1] or [5].

For this article we are motivated by papers [2] and [6]. In [2] the authors proved, by probabilistic methods, the following

Proposition 1. If a bounded sequence (c_{n}) behaves regularly with index −β, β ⩾ 0, then

exp(−x) ∑_{n=0}^{∞} \frac{c_{n}}{n!} x^{n} ∼ c_{[x]}, x → ∞.

In [6] we extend this proposition to the following

AMS Subject Classification: 26 A 12

Proposition 2. If exp P_{p}(x) = ∑_{n=0}^{∞} a_{n}x^{n}, where P_{p}(x) = b_{p}x^{p} + ⋯; b_{p} > 0, is a polynomial with (eventually) non-negative coefficients, then

exp(−P_{p}(x)) ∑_{n=0}^{∞} c_{n}a_{n}x^{n} ∼ (pb_{p})^{β}c_{[x^{p}]}, x → ∞; c_{0} = 1;

for any regularly varying sequence (c_{n}) of an arbitrary index β ∈ R.

Here we are going to show similar (and even more precise) asymptotic relations take place for segments of exponential series cited above.

Results

At the beginning we shall formulate a rather global proposition, showing how the structure of a given power series segment influence the behaviour of another one which involves regularly varying sequences. Namely, let us define

S(λ, x) := ∑_{n⩽λn(x)} a_{n}x^{n}, a_{n} ⩾ 0, n ∈ N;

where n(x) increases to infinity with x, and an operator T acting on S:

T S(λ, x) := ∑_{n⩽λn(x)} c_{n}a_{n}x^{n}, n ∈ N,

where (c_{n})_{n∈N} is a regularly varying sequence of index α ∈ R.

Theorem A. If there exist f, g_{1}, g_{2} : R^{+} → R^{+}, b_{1} : (0, 1) → R^{+}, b_{2} : (1, ∞) → R^{+}, and

lim_{x→∞} \frac{ln n(x)}{g_{i}(x)} = 0,  i = 1, 2,  (A1)

such that

\frac{S(λ, x)}{f(x)} = \{ \begin{matrix}O(e^{−b_{1}(λ)g_{1}(x)}), & 0 < λ < 1, \\ A + O(e^{−b_{2}(λ)g_{2}(x)}), & λ > 1, \end{matrix}  x → ∞, A ∈ R^{+}, (A2)

then

\frac{T S(λ, x)}{f(x)} = \{ \begin{matrix}o(c_{[n(x)]}), & 0 < λ < 1, \\ Ac_{[n(x)]}(1 + o(1)), & λ > 1, \end{matrix} x → ∞.

Proof. We shall use the following well-known properties of regularly varying functions ([5], pp. 19, 20; [1], p. 52):

For α > 0,

S_{1} : sup_{t⩽y} t^{α}L(t) = y^{α}L(y)(1 + o(1)); inf_{t⩾y} t^{α}L(t) = y^{α}L(y)(1 + o(1)), y → ∞;

S_{2} : inf_{t⩽y} t^{−α}L(t) = y^{−α}L(y)(1 + o(1)); sup_{t⩾y} t^{−α}L(t) = y^{−α}L(y)(1 + o(1)), y → ∞;

S_{3} : c_{[λy]} ∼ c_{[λ[y]]} ∼ λ^{α}c_{[y]}, y → ∞, λ ∈ R^{+}, α ∈ R.

In the sequel we consider a sequence (c_{n}) generated by some regularly varying function x^{α}L(x), α ∈ R, i.e.

c_{n} = n^{α}L(n), n ∈ N; c_{0} = 1; c_{[y]} = [y]^{α}L([y]).

Let α and λ (α ∈ R, 0 < λ < 1) be fixed numbers; then

\frac{T S(λ, x)}{c_{[n(x)]}f(x)} = \frac{1}{f(x)} ∑_{n⩽λn(x)} (\frac{n}{[n(x)]})^{α−1} (\frac{nL(n)}{[n(x)]L([n(x)])}) a_{n}x^{n}

⩽ \frac{1}{f(x)} sup_{n⩽λn(x)} (\frac{n}{[n(x)]})^{α−1} sup_{n⩽λn(x)} \frac{nL(n)}{[n(x)]L([n(x)])} ∑_{n⩽λn(x)} a_{n}x^{n}

= \frac{1}{f(x)}O([n(x)]^{|α|+1})O (\frac{[λn(x)]L([λn(x)])}{[n(x)]L([n(x)])}) O(e^{−b_{1}(λ)g_{1}(x)})

= O(n(x)^{|α|+1}e^{−b_{1}(λ)g_{1}(x)}) = O(exp(−b_{1}(λ)g_{1}(x)) (1 + O (\frac{ln n(x)}{g_{1}(x)}))

= o(1), x → ∞

Hence, the first assertion of Theorem A is proved.

For the second one, let λ and ε (λ > 1, 0 < ε < min(1/2, λ − 1)) be fixed. We get

\frac{T S(λ, x)}{c_{[n(x)]}f(x)}

= \frac{1}{c_{[n(x)]}f(x)}( ∑_{n⩽(1−ε)n(x)} + ∑_{(1−ε)n(x)<n⩽(1+ε)n(x)} + ∑_{(1+ε)n(x)<n⩽λn(x)})a_{n}c_{n}x^{n}
= T_{1} + T_{2} + T_{3}.

According to the former argument (λ = 1 − ε < 1),

T_{1} = o(1), x → ∞. (A3)

Analogously,

T_{3} ⩽ \frac{1}{f(x)} sup_{n⩽λn(x)} (\frac{c_{n}}{c_{[n(x)]}} ) ∑_{(1+ε)n(x)<n⩽λn(x)} a_{n}x^{n}

= \frac{1}{f(x)}O(n(x)^{|α|+1})(S(λ, x) − S(1 + ε, x))

= O(n(x)^{|α|+1})O(exp(−g_{2}(x) min(b_{2}(1 + ε), b_{2}(λ)))) = o(1), x → ∞. (A4)

To estimate T_{2}, suppose for the moment that index α of (c_{n}) is positive. Then, since 0 < ε < 1/2, using properties S_{1} and S_{3}, we obtain:

sup_{n⩽(1+ε)n(x)} c_{n} = c_{[(1+ε)n(x)]}(1 + o(1)) = c_{[n(x)]}(1 + ε)^{α}(1 + o(1))
= c_{[n(x)]}(1 + εO(1) + o(1)), x → ∞;
inf_{n>(1−ε)n(x)} c_{n} = c_{[(1−ε)n(x)]}(1 + o(1)) = c_{[n(x)]}(1 − ε)^{α}(1 + o(1))
= c_{[n(x)]}(1 + εO(1) + o(1)), x → ∞.

Therefore,

T_{2} ⩽ \frac{1}{f(x)c_{n(x)}} sup_{n⩽(1+ε)n(x)} c_{n} ∑_{(1−ε)n(x)<n⩽(1+ε)n(x)} a_{n}x^{n}

= \frac{1}{f(x)}(1 + εO(1) + o(1))(S(1 + ε, x) − S(1 − ε, x))

= (1 + εO(1) + o(1))(A + o(1)) = A + εO(1) + o(1), x → ∞, (A5)

and, similarly,

T_{2} ⩾ \frac{1}{f(x)c_{[n(x)]}} inf_{n>(1−ε)n(x)} c_{n} · (S(1 + ε, x) − S(1 − ε, x))
= A + εO(1) + o(1),  x → ∞.  (A6)

Since the constants in O(1) do not depend on ε and ε can be arbitrarily small, from (A5) and (A6) we conclude that T_{2} ∼ A, x → ∞; this, together with (A3) and (A4), gives the proof of Theorem A for α > 0.

For α < 0 we deduce the proof similarly, using properties S_{2} and S_{3}.

If α = 0, note that (nL(n)) is of index 1, hence

T_{2} = \frac{1}{f(x)L([n(x)])} ∑_{(1−ε)n(x)<n⩽(1+ε)n(x)} \frac{1}{n} · nL(n)a_{n}x^{n}

⩽ \frac{1}{[(1 − ε)n(x)] + 1}[(1 + ε)n(x)](A + o(1)),

and

T_{2} ⩾ \frac{1}{[(1 + ε)n(x)]}[(1 − ε)n(x)](A + o(1)),

which shows that in this case also T_{2} ∼ A, x → ∞, and the proof is over.

Investigation of possible relationship between S(λ, x), n(x) and f(x) satisfying conditions of Theorem A is the subject of our next article. Here we just show that the class of such functions is not empty, i.e. applying results of Theorem A we prove the following

Theorem B1. For any regularly varying sequence (c_{n})_{n∈N}, c_{0} = 1, of arbitrary index α ∈ R,

e^{−x} ∑_{n⩽λx} c_{n} \frac{x^{n}}{n!} ∼ \{ \begin{matrix}o(c_{[x]}), & 0 < λ < 1, \\ \frac{1}{2}c_{[x]}, & λ = 1, \\ c_{[x]}, & λ > 1, \end{matrix} x → ∞.

In the neighbourhood of λ = 1 we prove more precisely:

Segments of exponential series and regularly varying sequences 57

Theorem B2.

e^{−x} ∑_{n⩽x+h(x)} c_{n} \frac{x^{n}}{n!} ∼ (\frac{1}{2} + \frac{1}{\sqrt{π}} Erf(b/\sqrt{2})) c_{[x]}, x → ∞,

where h(x) := b\sqrt{x}(1 + o(1)), x → ∞; b ∈ R; Erf y := ∫_{0}^{y} e^{−t^{2}} dt.

Proof. According to the premises from Theorem A, the proof of cited theorems depends on asymptotic behaviour of the sum ∑_{k⩽n} \frac{x^{k}}{k!} , n = n(x) → ∞. Therefore, we derive its integral representation which is more easy to estimate.

S(n, x) := ∑_{k⩽n} \frac{x^{k}}{k!} = \frac{x^{n+1}}{n!} ∑_{k⩽n} \binom{n}{k}\frac{k!}{x^{k+1}} = \frac{x^{n+1}}{n!} ∑_{k⩽n} \binom{n}{k} ∫_{0}^{∞} e^{−xt}t^{k} dt,

i.e.

S(n, x) = \frac{x^{n+1}}{n!} ∫_{0}^{∞} e^{−xt}(1 + t)^{n} dt.  (B0)

For n = [λx] we obtain

e^{−x} \frac{x^{n+1}}{n!} ∼ \frac{x}{\sqrt{2πn}}e^{n ln x−n ln n+n−x} = \frac{x}{\sqrt{2πn}}e^{−x( \frac{n}{x} ln \frac{n}{x} +1− \frac{n}{x})}, x → ∞. (B1)

But

λ − \frac{1}{x} = \frac{λx − 1}{x} < \frac{n}{x} = \frac{[λx]}{x} ⩽ \frac{λx}{x} = λ,

i.e. \frac{n}{x} = λ − \frac{θ}{x}, θ ∈ [0, 1). Therefore,

\frac{n}{x} ln \frac{n}{x} + 1 − \frac{n}{x} = λ ln λ + 1 − λ − \frac{θ}{x} ln λ + 0 (\frac{1}{x^{2}} ) , x → ∞,

i.e. ((B1))

e^{−x} \frac{x^{n+1}}{n!} = O(\sqrt{x} e^{−(λ ln λ+1−λ)x}),  n = [λx], λ ∈ R^{+}, x → ∞. (B2)

Since ln(1 + t) < t, t ∈ R^{+}, for 0 < λ < 1, n = [λx], we get

∫_{0}^{∞} e^{−xt}(1 + t)^{n} dt < ∫_{0}^{∞} e^{−(x−n)} dt = \frac{1}{x − n} ∼ \frac{1}{x(1 − λ)}, x → ∞.

Along with (B2) this gives the estimate

e^{−x}S([λx], x) = e^{−x} ∑_{k⩽λx} \frac{x^{k}}{k!} = O(e^{−(λ ln λ+1−λ)x}),  λ ∈ (0, 1), x → ∞. (B3)

For l > 1, change of variable 1 + t ↦ t gives ∫_{0}^{∞} e^{−xt}(1 + t)^{n} dt = e^{x} ∫_{1}^{∞} e^{−xt}t^{n} dt = e^{x}(∫_{0}^{∞} − ∫_{0}^{1})e^{−xt}t^{n} dt = e^{x}(I_{1} + I_{2}), and, obviously, I_{1} = \frac{n!}{x^{n+1}} , |I_{2}| = O (\frac{e^{−x}}{x} ), which, together with (B0) and (B2) gives

e^{−x}S([λx], x) = 1 + O(e^{−x(λ ln λ+1−λ)}),  λ ∈ (1, ∞), x → ∞.  (B4)

Note that b(λ) := λ ln λ − λ + 1 is non-negative and convex on λ ∈ (0, +∞), b(0^{+}) = 1, b(1) = 0, b(λ) > 0 for λ ≠ 1, and

b(λ) > \{ \begin{matrix}\frac{1}{2}(1 − λ)^{2}, & λ ∈ (0, 1), \\ \frac{1}{2} ln^{2} λ, & λ ∈ (1, +∞). \end{matrix}

Comparing (B3) and (B4) with assertions from Theorem A, we see that conditions (A1), (A2) are satisfied with

f(x) = e^{x}, n(x) = g_{1}(x) = g_{2}(x) = x, b_{1}(λ) = \frac{(1 − λ)^{2}}{2} , b_{2}(λ) = \frac{ln^{2} λ}{2}, A = 1,

from which follows the validity of the first and the third assertion from Theorem B1. To prove Theorem B2, chnage variable in (B0): 1 + t ↦ \frac{n}{x}(1 + \frac{t}{\sqrt{n}}). We get:

e^{−x}S(n, x) = \frac{\sqrt{n}}{x} · \frac{x^{n+1}}{n!} · e^{−x} ∫_{\sqrt{n}( \frac{x}{n} −1)}^{∞} e^{−n+x} (\frac{n}{x})^{n} e^{−\sqrt{n} t+n ln(1+ \frac{t}{\sqrt{n}})} dt

= \frac{\sqrt{n}}{n!}n^{n}e^{−n}(∫_{0}^{∞} + ∫_{\sqrt{n}( \frac{x}{n} −1)}^{0})e^{−(\sqrt{n} t−n ln(1+ \frac{t}{\sqrt{n}}))} dt.  (B5)

Denote the first integral in (B5) by J_{1} and the second by J_{2} and let

g(n, t) := \sqrt{n} t − n ln (1 + \frac{t}{\sqrt{n}}) , t ⩾ 0, n ∈ N.

From the facts:

I: g(n, t) is monotone increasing on n.

Proof. 0 ⩽ ∫_{1}^{1+t/\sqrt{n}} \frac{1}{2s}(\sqrt{s} − \frac{1}{\sqrt{s}})^{2} ds = \frac{1}{2}(s − \frac{1}{s}) − ln s|_{1}^{1+t/\sqrt{n}} = g′_{n}(n, t);

II: lim_{n→∞} g(n, t) = \frac{t^{2}}{2} , t ∈ R^{+};

III: For n = [x + h(x)] follows \sqrt{n}(\frac{x}{n} − 1) → −b, x → ∞; using Lebesgue’s theorem of dominated convergence, we have:

J_{1} → ∫_{0}^{∞} e^{−t^{2}/2} dt = \sqrt{\frac{π}{2}} ;  (B6)

J_{2} → ∫_{−b}^{0} e^{−t^{2}/2} dt = \sqrt{2} ∫_{0}^{b/\sqrt{2}} e^{−t^{2}} dt = \sqrt{2} Erf(b/\sqrt{2}).  (B7)

Since

\frac{\sqrt{n}}{n!} · n^{n}e^{−n} → \frac{1}{\sqrt{2π}} , n → ∞,

from (B5), (B6), (B7) and Theorem A, the assertion of Theorem B2 follows. Putting b = 0 we obtain the sedond proposition from Theorem B1.

REFERENCES

[1] Bingham, N. H., Goldie, C. M., Teugels, J. L., Rgeular Variation, Camb. Univ. Press, 1989.

[2] Bingham, N. H., Hawkes, J., Some limit theorems for occupation times, London Math. Soc. Lecture Notes, 79, CUP, 1983, 46–62.

[3] Bojanic, R., Seneta, E., A unified theory of regularly varying sequences, MZ 134 (1973).

[4] Karamata, J., Sur un mode de croissance reguliere. Theoremes fondamenteaux, Bull. Soc. Math. France 61 (1933).

[5] Seneta, E., Functions of Regular Variations, Lecture Notes in Math. 508, Springer, 1976.

[6] Simic, S., Entire functions of finite order and regularly varying sequences, MZ (to appear)

(received 30.12.1998.)

Baba Višnjina 34, 11000 Beograd, Yugoslavia E-mail: ssimic@turing.mi.sanu.ac.yu