Some inequalities related to two expansions of (1+1/x)x$(1+1/x)^{x}$
Abstrak
AbstractWe prove the following theorem: Let (1+1x)x=e(1−∑k=1∞bk(1+x)k)=e(1−∑k=1∞dk(1112+x)k),σm(x)=∑k=1mbk(1+x)kandSm(x)=∑k=1mdk(1112+x)k. $$\begin{aligned}& \biggl(1+\frac{1}{x} \biggr)^{x}=e \Biggl(1- \sum _{k=1}^{\infty}\frac{b_{k}}{ (1+x )^{k}} \Biggr)=e \Biggl(1-\sum _{k=1}^{\infty}\frac{d_{k}}{ (\frac{11}{12}+x )^{k}} \Biggr), \\& \sigma_{m}(x)=\sum_{k=1}^{m} \frac{b_{k}}{ (1+x )^{k}} \quad\mbox{and}\quad S_{m}(x)=\sum_{k=1}^{m} \frac{d_{k}}{ (\frac{11}{12}+x )^{k}}. \end{aligned}$$ (1)If m≥6$m\geq6$ is even, we have Sm(x)>σm(x)$S_{m}(x)>\sigma_{m}(x)$ for all x>0$x>0$.(2)If m≥7$m\geq7$ is odd, we have Sm(x)>σm(x)$S_{m}(x)>\sigma_{m}(x)$ for all x>1$x>1$. This provides an intuitive explanation for the main result in Mortici and Hu (On an infinite series for (1+1/x)x$(1+ 1/x)^{x}$, 2014, arXiv:1406.7814 [math.CA]).
Topik & Kata Kunci
Penulis (2)
Bijun Ren
X. Li
Akses Cepat
- Tahun Terbit
- 2015
- Bahasa
- en
- Total Sitasi
- 1×
- Sumber Database
- Semantic Scholar
- DOI
- 10.1186/S13660-015-0928-5
- Akses
- Open Access ✓