设 A=(1+2/3^3)(1+2/5^3)(1+2/7^3)…(1+2/2019^3)，B=√(3/2)，比较 A 与 B 的大小

myyour

比较两个数的大小。

myyour

我想了两天都想不出，请老师们帮帮忙，谢谢！

luyuanhong

楼上 永远 转发的解答很好！已收藏。

uk702

luyuanhong 发表于 2020-11-6 22:02
楼上 永远 转发的解答很好！已收藏。

我怎么看不到？能否给个链接？谢谢。

uk702

如图所示。

luyuanhong

uk702 发表于 2020-11-8 06:42
我怎么看不到？能否给个链接？谢谢。

下面是网友 永远 转发的解答：

uk702

回复：@Nicolas2050

多谢指正！请再检查。

王守恩

电脑是这样出来的，有兴趣的不妨欣赏一下。

1，\(\displaystyle\prod_{k=1}^{1009}\bigg(1+\frac{2}{(2k+1)^3}\bigg)\)
   1.1060942341644551225679854213644773065200754073846

2，\(\displaystyle\prod_{k=1}^{\infty}\bigg(1+\frac{2}{(2k+1)^3}\bigg)\)
   1.10609436970190620636951610859849126125138224376556  

3，\(\displaystyle\prod_{k=1}^{\infty}\bigg(1+\frac{2}{(2k+1)^3}\bigg)\)// FullSimplify
   \[Pi]^(3/2)/(2 (-1 + 2^(2/3)) Gamma[1/2 (3 - (-2)^(1/3))]
   Gamma[3/2 + (-(1/2))^(2/3)] Gamma[-(1/2) + 1/2^(2/3)])

4，\(\displaystyle\prod_{k=1}^{\infty}\bigg(1+\frac{1}{(k+n)^2}\bigg)\), {n, 0, 9}]
   Sinh[\[Pi]]/\[Pi],
   Sinh[\[Pi]]/(2 \[Pi]),
   (2 Sinh[\[Pi]])/(5 \[Pi]),
   (9 Sinh[\[Pi]])/(25 \[Pi]),
   (144 Sinh[\[Pi]])/(425 \[Pi]),
   (72 Sinh[\[Pi]])/(221 \[Pi]),
   (2592 Sinh[\[Pi]])/(8177 \[Pi]),
   (63504 Sinh[\[Pi]])/(204425 \[Pi]),
   (4064256 Sinh[\[Pi]])/(13287625 \[Pi]),
   (164602368 Sinh[\[Pi]])/(544792625 \[Pi])

5，\(\displaystyle\prod_{k=1}^{\infty}\bigg(1+\frac{1}{(k+n)^3}\bigg)\),{n, 0, 8}] // FullSimplify
   Cosh[(Sqrt[3] \[Pi])/2]/\[Pi],
   Cosh[(Sqrt[3] \[Pi])/2]/(2 \[Pi]),
   (4 Cosh[(Sqrt[3] \[Pi])/2])/(9 \[Pi]),
   (3 Cosh[(Sqrt[3] \[Pi])/2])/(7 \[Pi]),
   (192 Cosh[(Sqrt[3] \[Pi])/2])/(455 \[Pi]),
   (800 Cosh[(Sqrt[3] \[Pi])/2])/(1911 \[Pi]),
   (57600 Cosh[(Sqrt[3] \[Pi])/2])/(138229 \[Pi]),
   (7200 Cosh[(Sqrt[3] \[Pi])/2])/(17329 \[Pi]),
   (409600 Cosh[(Sqrt[3] \[Pi])/2])/(987753 \[Pi])

6，\(\displaystyle\prod_{k=1}^{\infty}\bigg(1+\frac{2}{(k+n)^2}\bigg)\),{n, 0, 6}] // FullSimplify
   Sinh[Sqrt[2] \[Pi]]/(Sqrt[2] \[Pi]),
   Sinh[Sqrt[2] \[Pi]]/(3 Sqrt[2] \[Pi]),
   (Sqrt[2] Sinh[Sqrt[2] \[Pi]])/(9 \[Pi]),  
   (Sqrt[2] Sinh[Sqrt[2] \[Pi]])/(11 \[Pi]),
   (8 Sqrt[2] Sinh[Sqrt[2] \[Pi]])/(99 \[Pi]),
   (200 Sqrt[2] Sinh[Sqrt[2] \[Pi]])/(2673 \[Pi]),
   (400 Sqrt[2] Sinh[Sqrt[2] \[Pi]])/(5643 \[Pi])

7，\(\displaystyle\prod_{k=1}^{\infty}\bigg(1+\frac{2}{(2k+n)^3}\bigg)\),{n, 0, 4}] // FullSimplify
   2^(2/3)/(Gamma[1/2^(2/3)] Gamma[1/2 (2 - (-2)^(1/3))] Gamma[1 + (-(1/2))^(2/3)]),
   \[Pi]^(3/2)/(2 (-1 + 2^(2/3)) Gamma[1/2 (3 - (-2)^(1/3))]
Gamma[3/2 + (-(1/2))^(2/3)] Gamma[-(1/2) + 1/2^(2/3)]),
   (4 2^(2/3))/(5 Gamma[1/2^(2/3)] Gamma[1/2 (2 - (-2)^(1/3))] Gamma[1 + (-(1/2))^(2/3)]),
   -((27 \[Pi]^(3/2))/( 8 (1 + 2^(1/3) - 3 2^(2/3)) Gamma[1/2 (5 - (-2)^(1/3))]
Gamma[ 5/2 + (-(1/2))^(2/3)] Gamma[-(1/2) + 1/2^(2/3)])),
   (128 2^(2/3))/(165 Gamma[1/2^(2/3)] Gamma[1/2 (2 - (-2)^(1/3))] Gamma[1 + (-(1/2))^(2/3)])

王守恩

luyuanhong 发表于 2020-11-8 07:24
下面是网友 永远 转发的解答：

谢谢永远！整理一下。
       \(\displaystyle\prod_{k=1}^{1009}\big(1+\frac{2}{(2k+1)^3}\big)\)
\(\displaystyle=\)e^\(\displaystyle\bigg(\sum_{k=1}^{1009}\ \ln\big(1+\frac{2}{(2k+1)^3}\big)\bigg)\)
\(\displaystyle<\)e^\(\displaystyle\bigg(\sum_{k=1}^{1009}\ \big(\frac{2}{(2k+1)^3}\big)\bigg)\)
\(\displaystyle<\)e^\(\displaystyle\bigg(\sum_{k=1}^{1009}\ \big(\frac{2}{2k(2k+1)(2k+2)}\big)\bigg)\)
\(\displaystyle=\)e^\(\displaystyle\bigg(\sum_{k=1}^{1009}\ \big(\frac{1}{2k(2k+1)}-\frac{1}{(2k+1)(2k+2)}\big)\bigg)\)
\(\displaystyle<\)e^\(\displaystyle\ \big(\frac{1}{2×3}\big)<\sqrt{\frac{3}{2}}\)