The asymptotic equivalence of LR, Wald and score tests Suppose that $Y_1, \ldots, Y_{n}$ are iid from a Bernoulli distribution with parameter $p$ and consider $H_0 : p = p_0\,.$ The test statistics are 
$$ T_W = \frac{n ({\widehat p} - p_0)^2}{{\widehat p} (1 - {\widehat p})}\,, \:\:\:\: T_S = \frac{n ({\widehat p} - p_0)^2}{p_0 (1 - p_0)} \,,$$
and $ T_{LR} = -2 [ X \log(p_0/{\widehat p}) + (n-X) \log[(1 - p_0)/(1 -\widehat p)] ], $
where $X = \sum_{i=1}^{n} Y_i\,.$ 
I want to Show that $T_W, T_S$ and $T_{LR}$ are asymptotically equivalent under $H_0$ by showing that their differences converge to $0$ in probability.
I showed that $T_W $ is  asymptotically equivalent to $T_S$ as follows
We have 
\begin{align*}
T_W-T_S &=n ({\widehat p} - p_0)^2 [\frac{1}{{\widehat p} (1 - {\widehat p})}-  \frac{1}{p_0 (1 - p_0)}]\\
&= p_0 (1 - p_0) \left(  \frac{\sqrt{n} ({\widehat p} - p_0)}{\sqrt{p_0 (1 - p_0) }}\right)^{2}[\frac{1}{{\widehat p} (1 - {\widehat p})}-  \frac{1}{p_0 (1 - p_0)}]
\end{align*}
From the asymptotic normality of the MLE, we have that
$$\left(  \frac{\sqrt{n} ({\widehat p} - p_0)}{\sqrt{p_0 (1 - p_0) }}\right)^{2} \overset{d}{\rightarrow} \mathcal{N}(0,1)^{2}=\chi^{2}(1),$$
and from the consistency of the MLE we have that 
$$  {\widehat p} \overset{p}{\rightarrow} p_0.  $$
From Slutsky’s theorem and the continuous mapping theorem (CMT) we have
$$ [\frac{1}{{\widehat p} (1 - {\widehat p})}-  \frac{1}{p_0 (1 - p_0)}] \overset{p}{\rightarrow} 0.$$
This implies that
\begin{align*}
 T_W-T_S  & \overset{d}{\rightarrow} 0\\
 T_W-T_S  & \overset{p}{\rightarrow} 0.
\end{align*}
But I did not succeed at showing the same conclusion for $T_W$ and $T_{LR}$. Can someone help?

Use the Taylor expansion of $T_{LR}$ around $p_0$:
\begin{align}
T_{LR}&=-2n\left[\hat p \ln\left(\frac{p_0}{\hat p}\right) + (1-\hat p) \ln\left(\frac{1-p_0}{1-\hat p}\right)\right] \\
&=\frac{n}{p_0(1-p_0)}(\hat p-p_0)^2+O_p(|\hat p-p_0|^3) \\[1em]
&=T_S+o_p(1)=T_W+o_p(1).
\end{align}
