Abstract
For any two real numbers α > 0 and β > −α , we show that the best constants a and b (the smallest a and the largest b) such that the inequalities ∞ 1 1 < k=∑n+1 (α − k 1 + )k−β1< 2α n + a 2α n + b hold for every n > 1 are a = (α+1β − S(α, β ) )−1 − 2α and b = α + 2β, where S(α, β ) = n∑=∞1(α−n1+)n−β1 . In particular, we recover the main result of [6] and answer a question, stated in [6], about the Gregory-Leibniz series ∑∞ (−2n1−)n−11 . More precisely, we show that the best constants n=1 c and d such that the inequalities ∞ 1 1 < k=∑n+1(−2k1−)k−11 < 4n + d 4n + c 4 hold for every n > 1 are c = − 4 and d = 0.
| Original language | English |
|---|---|
| Pages (from-to) | 83-91 |
| Number of pages | 9 |
| Journal | Mathematical Inequalities and Applications |
| Volume | 26 |
| Issue number | 1 |
| DOIs | |
| State | Published - Jan 2023 |
Bibliographical note
Publisher Copyright:© 2023 Element D.O.O.. All rights reserved.
Keywords
- Alternating series
- estimate of the remainder of a series
- hypergeometric series
ASJC Scopus subject areas
- General Mathematics
- Applied Mathematics
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