Why this approximation not complete with me?

Question

consider the sequence $$ M \sim \sqrt{2\pi} (\frac{n}{e})^n \sqrt[6]{8n^3+4n^2+n+\frac{1}{30}-\frac{1}{K_1n+K_2+\frac{T_1}{n}+\frac{T_2}{n^2}+\frac{T_3}{n^3}}} $$

prove that $$ M \sim \sqrt{\pi} (\frac{n}{e})^n \sqrt[6]{8n^3+4n^2+n+\frac{1}{30}-U_{n}},~~ n\geq1$$ where

$$ U_n= \frac{1}{\frac{240n}{11}+\frac{9480}{847}+\frac{919466}{65219n}+\frac{1455925}{5021863n^2}+\frac{639130140092}{92804028n^3}} $$ i found the value of $k_1,k_2,T_1 $ by measure the accuracy of this approximation by define the sequence by relation $$ R_n=\ln n!-\ln \sqrt{2\pi}-n \ln n +n -\frac{1}{6}\ln\left(8n^3+4n^2+n+\frac{1}{30}-\frac{1}{K_1n+K_2}\right) $$ and get $k_1,k_2$

then defined it again and get $T_1$ by developing $R_n-R_{n+1}$ in power series in $\frac{1}{n}$ and used Taylor in mathematica prog mathematica prog, and used same method to get $T_2,T_3$ but not get with me as they are in $U_n$, any one can know the reason

Answer 1

Starting from the Stirling approximation $$ \ln n!\sim n\ln n-n+{\tfrac {1}{2}}\ln(2\pi n)+{\frac {1}{12n}}-{\frac {1}{360n^{3}}}+{\frac {1}{1260n^{5}}}-{\frac {1}{1680n^{7}}}+\cdots $$ we get by simple manipulation of truncated Taylor series \begin{align} &\ln n!-n\ln n-n-\frac12\ln(\pi)\sim\frac12\ln(2n)+{\frac {1}{12n}}-{\frac {1}{360n^{3}}}+{\frac {1}{1260n^{5}}}-{\frac {1}{1680n^{7}}}+\cdots \\ &=\frac16\ln\left(8n^3\exp\left({\frac {1}{2n}}-{\frac {1}{60n^{3}}}+{\frac {1}{210n^{5}}}-{\frac {1}{280n^{7}}}+\cdots\right)\right) \\ &=\frac16\ln\left(8n^3 + 4n^2 + n + \frac1{30} - \frac{11}{240n} + \frac{79}{3360n^2} + \frac{3539}{201600n^3} - \frac{9511}{403200n^4} - \frac{10051}{716800n^5} + \cdots\right) \\ &=\frac16\ln\left(8n^3 + 4n^2 + n + \frac1{30} - \frac1{\frac{240n}{11} + \frac{9480}{847} + \frac{919466}{65219n} + \frac{1455925}{5021863n^2} - \frac{639130140029}{92804028240n^3} + \cdots }\right) \end{align}

Up to the last term, all other coefficients are the same as you got.

Answer 2

@luzl thanks for your help i know this
but i need to find this differenceby this way by taylor to find T_2 usch that i find $k1,k_2,T_1$ as in this seq ,not but $T_2$ not get as it's value in the top seq (n>1)i used (n from 0 to 7) \begin{eqnarray*} \frac{1}{n} \log[1 + n] -1 -\frac{1}{6} \log[ 8 (\frac{1}{n})^3 + 4 (\frac{1}{n})^2 + \frac{1}{n}+ \frac{1}{30} - \frac{1}{(\frac{240}{11} \frac{1}{n} + \frac{9480}{847}+\frac{919466}{65219\frac{1}{n}} +\frac{T_2}{(\frac{1}{n})^2}))}] +\frac{1}{6} \log[8 (\frac{1}{n} + 1)^3 +4 (\frac{1}{n}+1)^2 +(\frac{1}{n} + 1) +\frac{1}{30} -\frac{1}{(\frac{240}{11} (\frac{1}{n} + 1) + \frac{9480}{847}+\frac{919466}{(65219(\frac{1}{n}+1)}+\frac{T_2}{(\frac{1}{n}+1)^2})}] \end{eqnarray*}