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Assume $n\ge 2$, so the residual sum of squares is positive almost surely. Write $\overline{Y}=n^{-1}{\sum }_i{Y}_i$ and $S={\sum }_i(Y_i-\overline{Y}{)}^2$. The log-likelihood isThe decomposition ${\sum }_i(Y_i-\mu {)}^2=S+n(\overline{Y}-\mu {)}^2$ shows that, at every fixed positive variance, its constrained maximum-likelihood estimate of the mean is $\overline{Y}$. Substituting this mean and differentiating with respect to the variance givesThe denominator $n$ is required by maximum likelihood estimation; $S/(n-1)$ is the unbiased variance estimate, which is a different estimator.

Under the null variance ${\sigma }_0^2$, maximize over the same mean $\overline{Y}$. Subtracting the null maximum log-likelihood from the unrestricted maximum gives the likelihood-ratio test statisticBy Cochran's theorem, the residual sum of squares in a normal sample satisfies $U=S/{\sigma }_0^2\sim {\chi }_{n-1}^2$ under the null. Set $V=(U-n)/n$, so $U/n=1+V>0$. ThenThe normal mean is a nuisance parameter, but this null law does not depend on it.

A Bartlett correction removes the first-order discrepancy between the null mean of a regular likelihood-ratio test statistic and the degrees of freedom of its limiting chi-squared distribution. In the convention used here, if the limiting degrees of freedom are $d$ andthen $b$ is the Bartlett correction coefficient, also called the Bartlett correction factor, and the Bartlett-corrected likelihood-ratio statistic is $W_B=W/(1+b/n)$. Its null mean is $d+O(n^{-2})$. The finite-$n$ divisor is $1+b/n$, rather than $b$ itself. There is one restriction on the variance here, so $d=1$.

Expand the logarithm around $V=0$ and use the permitted termwise asymptotic integration:Using the supplied raw moments of the chi-squared distribution and expanding $(U-n{)}^r$ givesFor instance, $\mathbb{E}U=n-1$ and $\mathbb{E}{U}^2=(n-1)(n+1)$ give the first formula after subtracting $2n\mathbb{E}U-n^2$. The same binomial expansion, using the raw third and fourth moments, gives the other two formulas.

We must justify why higher moments cannot change the $n^{-1}$ coefficient. Represent $U={\sum }_{i=1}^{n-1}{Z}_i^2$ for independent standard normal variables, and put $T_i=Z_i^2-1$. Each $T_i$ is centered and has all absolute moments finite. With $H={\sum }_{i=1}^{n-1}{T}_i$, one has $V=(H-1)/n$. The supplied centered-sum bounds say, for each fixed integer $j\ge 2$,Also $\mathbb{E}H=0$ and $\mathbb{E}{H}^0=1$. Therefore, for every fixed $r\ge 2$, expanding $(H-1{)}^r$ givesThese are bounds on the absolute values of signed moments; the odd-moment estimate must not be misread as a bound on $\mathbb{E}|H{|}^j$. In particular, each fixed Taylor term of degree $r\ge 5$ has $n\mathbb{E}{V}^r=O(n^{-2})$ or smaller. Combined with the question's permission to integrate the asymptotic expansion term by term, this controls the higher-order contribution at the required order. It does not require a globally convergent Taylor series for every realized $V$.

Substituting the first three moment formulas now givesThus the Bartlett correction for a normal variance with unknown mean isIts null mean is $1+O(n^{-2})$, and the usual reference law is ${\chi }_1^2$. Keeping the estimated mean's loss of one residual degree of freedom, and using the maximum-likelihood variance divisor $n$, are both essential to the coefficient $11/6$.