3# 的方法就是大招。更进一步，似有
\[
a_n = \frac{2}{n} + \frac{2 Log}{3 n^2} + \frac{2 Log^ ...
I am very sorry! #7 楼的结果没能通过实际数据检验。
n = 100000;
a = 1.0`100; For; i++]; a1000000 = a; Print];
f := 2/n + (2/3 Log - c1)/(n^2); Solve == a]
g := 2/n + (2/3 Log - c2)/(n^2) + (2 Log Log - 2 Log)/(9 n^3); Solve == a]
解得
c1 = 4.513881773021752...
c2 = 4.514150738637968...
n = 1000000;
a = 1.0`100; For; i++]; a1000000 = a; Print];
f := 2/n + (2/3 Log - c1)/(n^2); Solve == a]
g := 2/n + (2/3 Log - c2)/(n^2) + (2 Log Log - 2 Log)/(9 n^3); Solve == a]
解得
c1 = 4.5139127439173691673462937...
c2 = 4.513952088988795669136132843492
故(猜想)
\[
a_n = \frac{2}{n} + \frac{2/3 Log - c}{n^2}+ o(n^{-2})
\]
其中 c 是一个与 \( a_1 \) 相关的常数，当\( a_1 = 1 \)时，\( c = 4.513... \)
根据3#的思想，得到$n->\infty,a(n)->2\sum _{k=1}^{\infty } \frac{1}{2 k-1} (\frac{1}{n})^{2 k-1}= 2 \tanh ^{-1}(\frac{1}{n})$,也就是原题极限为0. wayne 发表于 2023-1-15 09:18
根据3#的思想，得到$n->\infty,a(n)->2\sum _{k=1}^{\infty } \frac{1}{2 k-1} (\frac{1}{n})^{2 k-1}= 2 ...
你好，原极限结果等于2/3，2楼的程序可否能改进一下 本帖最后由 uk702 于 2023-1-15 10:40 编辑
\[
a_n = \frac{2}{n} + \frac{2/3 Log - c}{n^2} + \frac{ 2/9 Log^2 - 2(1 + 3 c)/ 9 Log + (2 + 6 c + 9 c^2)/18}{n^3}
\]
n = 100000;
a = 1.0`100; For; i++]; Print];
h = 2/n + (2/3 Log - c) / n^2 + ( 2/9 Log^2 - 2(1 + 3 c)/ 9 Log + (2 + 6 c + 9 c^2)/18 ) / n^3;
Solve == a]
解得 c = 4.513922317288945344462772324200553...
n = 1000000;
a = 1.0`100; For; i++]; Print];
h = 2/n + (2/3 Log - c) / n^2 + ( 2/9 Log^2 - 2(1 + 3 c)/ 9 Log + (2 + 6 c + 9 c^2)/18 ) / n^3;
Solve == a]
解得 c = 4.5139223177270656672830726730444912910404826975...
n = 10000000;
a = 1.0`100; For; i++]; Print];
h = 2/n + (2/3 Log - c) / n^2 + ( 2/9 Log^2 - 2(1 + 3 c)/ 9 Log + (2 + 6 c + 9 c^2)/18 ) / n^3;
Solve == a]
解得 c = 4.51392231774456470363880866511722354631999512440581836558125598...
这时a = 2.0000006231476527853498662133367261421598872948834435881757796... *10^-7
h = 2.00000062314765278534986621333672614215988729488344358817577963496395 *10^-7
注意到 h == a，上面的结果其实没有意义，要考察的是 c 对不同的 n 的稳定性，由于 c 的精度超过了 1/n，介于 1/n 和1/n^2 之间，故判断上述公式给出的 1/n^3 的系数正确，而 （未给出的） 1/n^4 的系数可能较大。
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