Why this approximation not complete with me?

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

• Where do you think the factor $\sqrt2$ vanishes to? Asymptotically, this is the difference between the two expressions. – LutzL Jul 11 at 20:36
• \begin{eqnarray*} M = n \Log[1 + \frac{1}{n} - 1 - \frac{1}{6} \Log[ (8 n^3 + 4 n^2 + n + \frac{1}{30 }- \frac{1}{(K1n + K2)}] + \frac{1}{6} \Log[(8 (n + 1)^3 + 4 (n + 1)^2 + (n + 1) + \frac{1}{30 } - \frac{1}{( K1 (n + 1) + K2))}] \end{eqnarray*} ,this the difference between the sequence ,by mathematica prog (taylor series) i get $k_1=\frac{240}{11}$,and $k_2=\frac{9480}{847}$ – researcher Jul 11 at 21:22
• What I mean is that in your first and also the last formula instead of $2\pi$ it has to be just $\pi$, the factor 2 is used under the 6th root. – LutzL Jul 11 at 21:25
• but $k_2$ not get with me as the value in the original sequence! – researcher Jul 11 at 21:28

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}
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) $$\begin{eqnarray*} \frac{1}{n} \log[1 + n] -1 -\frac{1}{6} \log[ 8 (\frac{1}{n})^3 + 4 (\frac{1}{n})^2 + \frac{1}{n}+ \frac{1}{30} - \frac{1}{(\frac{240}{11} \frac{1}{n} + \frac{9480}{847}+\frac{919466}{65219\frac{1}{n}} +\frac{T_2}{(\frac{1}{n})^2}))}] +\frac{1}{6} \log[8 (\frac{1}{n} + 1)^3 +4 (\frac{1}{n}+1)^2 +(\frac{1}{n} + 1) +\frac{1}{30} -\frac{1}{(\frac{240}{11} (\frac{1}{n} + 1) + \frac{9480}{847}+\frac{919466}{(65219(\frac{1}{n}+1)}+\frac{T_2}{(\frac{1}{n}+1)^2})}] \end{eqnarray*}$$