Likelihood Ratio Test for the Normal Distribution with unknown mean Let $(X_{1},...,X_{n})$ a $n$ sample of the law $N( \mu, \sigma^{2})$.
We assumed we don't know $\mu$ and $\sigma^{2}$.
Let $\mu_{0} \in \mathbb{R}$. 

Show that the Likelihood-ratio test for $\mu = \mu_{0}$ against $\mu
 \ne \mu_{0}$ is function of  $$ 1 + \frac{(\bar{X_{n}} - \mu_{0}
 )^{2}}{\sigma_{n}^{2}} $$ with $\sigma_{n}^{2}  = \sum_{i=1}^{n}
 \frac{(X_{i} - \bar{X_{n}})^{2}}{n}$

EDIT : I showed that the Likelihood-ratio test is
$$
\exp\left( \frac{n}{2 s_{n}^{2}(X)} (\bar{X_{n}} - \mu_{0} ) \right)
$$ 
with $s_{n}^{2}(X) = \sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n}$
But I don't know how to conclude.
Thanks and regards.

$L_{H_0} = (2\pi\sigma^2)^{-\frac{n}{2}}exp(-\frac{1}{2\sigma^2}\sum_{i=1}^n(X_i-\mu_0)^2)$
$supL_{H_1} = (2\pi\sigma^2)^{-\frac{n}{2}}exp(-\frac{1}{2\sigma^2}\sum_{i=1}^n(X_i-\bar{X})^2)$ ($\bar{X}$ is the maximum likelihood estimate for $\mu$ if $\mu$ is unknown). 
The problem is that you are treating $\sigma^2$ as given parameter instead of unknown. If you maximize the likelihood wrt $\sigma^2$ for ${H_0}$ you will obtain that the maximum likelihood estimate  for $\sigma^2$ is $\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n}$ (analogously for ${H_1}$ it is $\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n}$). 
After plugging in the estimate for $\sigma^2$ you will obtain:
$\sup L_{H_0} = (2\pi\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n})^{-\frac{n}{2}}exp\Bigg(-\frac{\sum_{i=1}^n(X_i-\mu_0)^2}{2{\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n}}}\Bigg)=\big(\frac{2\pi}{n}\sum_{i=1}^n(X_i-\mu_0)^2)^{-\frac{n}{2}}exp(-\frac{n}{2})$ 
and $\sup L_{H_1} = (2\pi\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n})^{-\frac{n}{2}}exp\Bigg(-\frac{\sum_{i=1}^n(X_i-\bar{X}^2)}{2{\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n}}}\Bigg) =\big(\frac{2\pi}{n}\sum_{i=1}^n(X_i-\bar{X})^2)^{-\frac{n}{2}}exp(-\frac{n}{2})$.
I.e. $\Lambda = \Bigg(\frac{\sum_{i=1}^n(X_i-\mu_0)^2}{\sum_{i=1}^n(X_i-\bar{X})^2}\Bigg)^{-\frac{n}{2}}$.
$\sum_{i=1}^n(X_i-\mu_0)^2 = \sum_{i=1}^n((X_i-\bar{X})+(\bar{X}-\mu_0))^2 =  \sum_{i=1}^n(X_i-\bar{X})^2+n(\bar{X}-\mu_0)^2$.
Plugging back into the equation for $\Lambda$ we obtain: $\Lambda = \Bigg(1+\frac{n(\bar{X}-\mu_0)^2}{\sum_{i=1}^n(X_i-\bar{X})^2}\Bigg)^{-\frac{n}{2}}$.
