$\sqrt n (\hat{\theta}-\theta) \ →_d \ N(0,\sigma^2）$ implies $plim \ \hat{\theta} = \theta $

Is this correct? If so, how can I prove this?

My proof is like this:

Since $\hat{\theta}→_{asy.} N(\theta, \frac{\sigma^2}{n}) $
I have $\lim_{n \to \infty} E(\hat{\theta})= \theta \ $ and $\lim_{n \to \infty} Var(\hat{\theta})=0 $.

Therefore, from convergence in quadratic mean, $plim \ \hat{\theta} = \theta $.

Is this proof correct?

Is it true that $\sqrt n (\hat{\theta}-\theta) \ \rightarrow_d \ N(0,\sigma^2)$ implies $\text{plim} \ \hat{\theta} = \theta $? If so, how can I prove this?

Attempted proof: My proof is like this:

Since $\hat{\theta} \rightarrow_{asy.} N(\theta, \frac{\sigma^2}{n})$ I have $\lim_{n \to \infty} E(\hat{\theta})= \theta \ $ and $\lim_{n \to \infty} Var(\hat{\theta})=0 $. Therefore, from convergence in quadratic mean, $\text{plim} \ \hat{\theta} = \theta $. Is this proof correct?

$\sqrt n (\hat{\theta}-\theta) \ →_d \ N(0,\sigma^2） $ implies $plim \ \hat{\theta} = \theta $
Is this correct? If so, how can I prove this?

My proof is like this.
Since $\hat{\theta}→_{asy.} N(\theta, \frac{\sigma^2}{n}) $
I have $\lim_{n \to \infty} E(\hat{\theta})= \theta \ $ and $\lim_{n \to \infty} Var(\hat{\theta})=0 $. Therefore, from convergence in quadratic mean, $plim \ \hat{\theta} = \theta $.

Is this proof correct?

$\sqrt n (\hat{\theta}-\theta) \ →_d \ N(0,\sigma^2）$ implies $plim \ \hat{\theta} = \theta $

Is this correct? If so, how can I prove this?

My proof is like this:

Since $\hat{\theta}→_{asy.} N(\theta, \frac{\sigma^2}{n}) $
I have $\lim_{n \to \infty} E(\hat{\theta})= \theta \ $ and $\lim_{n \to \infty} Var(\hat{\theta})=0 $.

Therefore, from convergence in quadratic mean, $plim \ \hat{\theta} = \theta $.

Is this proof correct?