Calculate the consistency of an Estimator I need to determine whether the following estimator $T$ is asymptotically unbiased and consistent for an i.i.d. sample of Gaussian distributions with $X_{i} \sim N(\mu, \sigma)$:
\begin{equation*}
T = \frac{1}{2} X_{1} + \frac{1}{2n} \sum\limits_{i = 2}^{n} X_{i}
\end{equation*}
Keep in mind that $n$ denotes the sample size.
I was able to figure out that the estimator $T$ is asymptotically unbiased.
First, I determined the expected value of the estimator.
\begin{align*}
    E[T] &= \frac{1}{2} \cdot E[X_{1}] + \frac{1}{2 \cdot n} \sum\limits_{i = 1}^{n} E[X_{i}] \\
    &= \frac{E[X]}{2} + \frac{1}{2 \cdot n} \cdot (n - 1) \cdot E[X] \\
    &= \frac{E[X]}{2} + \frac{(n - 1)}{2 \cdot n} \cdot E[X] \\
    &= \frac{\mu}{2} + \frac{(n - 1) \cdot \mu}{2 \cdot n} \\
    &= \frac{\mu}{2} + \frac{n \cdot \mu - \mu}{2 \cdot n} \\
\end{align*}
Since the expected value does not equal $\mu$, one can conclude that the estimator $T$ is biased.
However, if we calculate the estimator's bias $b(T)$ and check if it converges to 0, we can see that the estimator is asymptotically unbiased (the calculation of the limit was done using WolframAlpha):
\begin{equation*}
  b(T) = E[T] - \mu = \left(\frac{\mu}{2} + \frac{n \cdot \mu - \mu}{2 \cdot n} \right) - \mu
\end{equation*}
\begin{equation*}
  \text{lim}_{n \rightarrow +\infty}\left(\frac{\mu}{2} + \frac{n \cdot \mu - \mu}{2 \cdot n} - \mu \right) = 0
\end{equation*}
Unfortunately, I have not been able to find out whether the estimator $T$ is consistent.
From my understanding we can find out if a biased estimator is consistent by verifying if the mean squared error $MSE$ of the error approaches 0 when the sample size $n$ gets infinitely large.
In order to calculate the $MSE$, we need to calculate the variance $VAR$ of the estimator and then subtract the square of the bias $b$ from the variance $VAR$:
\begin{equation*}
\text{MSE}(T) = \text{VAR}(T) - b^{2}(T)
\end{equation*}
\begin{equation*}
\text{lim}_{n \rightarrow +\infty}\left(\text{MSE}(T)\right) = 0 \Rightarrow T \text{ is consistent}
\end{equation*}
The issue is that I am not able to correctly calculate the MSE.
I tried many approaches, but I could not figure out what's wrong.
My current approach is the following:
\begin{align*}
    \text{VAR}(T) &= \frac{1}{2^{2}} \cdot \text{VAR}(X_{1}) + \frac{1}{(2 \cdot n)^{2}} \sum\limits_{i = 1}^{n} \text{VAR}(X_{i}) \\
    &= \frac{1}{2^{2}} \cdot \text{VAR}(X) + \frac{1}{(2 \cdot n)^{2}} \cdot (n - 1) \cdot \text{VAR}(X) \\
    &= \frac{\sigma^{2}}{4} + \frac{(n - 1) \cdot \sigma^{2}}{4 \cdot n^{2}} \\
    &= \frac{\sigma^{2}}{4} + \frac{n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} \\
    &= \frac{n^{2} \cdot \sigma^{2}}{4 \cdot n^{2}} + \frac{n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} \\
    &= \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} \\
\end{align*}
The particular issue lies in finding the value for the square of the bias $b^{2}(T)$.
I tried many different approaches, but I could not find an equation which makes my calculation work.
Therefore, my issue is how can I find a sensible equation for $b^{2}(T)$?
Just for reference, here is my current approach:
See WolframAlpha for the expansion of the bias
\begin{align*}
b^{2}(T) &= \left( \frac{\sigma}{2} + \frac{n \cdot \sigma - \sigma}{2 \cdot n} - \sigma \right)^{2} \\
&= \frac{\sigma^{2}}{4} \\
\end{align*}
\begin{align*}
\text{MSE}(T) &= \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} - \frac{\sigma^{2}}{4}\\
&= \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - 2 \cdot \sigma^{2}}{4 \cdot n^{2}}
\end{align*}
\begin{equation*}
\text{lim}_{n \rightarrow +\infty} \left( \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - 2 \cdot \sigma^{2}}{4 \cdot n^{2}} \right) = \frac{\sigma^{2}}{4}
\end{equation*}
Thank you for your help! Grazie mille!

By definition, a consistent estimator converges in probability to a constant as the sample grows larger.
To be explicit, let's subscript $T$ with the sample size.  Note that
$$\operatorname{Var}(T_n) = \operatorname{Var}\left(\frac{X_1}{2}\right) + \operatorname{Var}\left(\frac{1}{2n}\sum_{i=2}^n X_i\right) \ge \operatorname{Var}\left(\frac{X_1}{2}\right) = \frac{\sigma^2}{4}.$$
Because $T_n,$ being a linear combination of independent Normal variables, has a Normal distribution, it cannot possibly converge to a constant and therefore is not consistent.
One quick rigorous proof is to suppose it does converge in probability to a number $\theta$ and then observe that $\Pr(|T_n-\theta|\ge \sigma) \ge \Phi(1)-\Phi(-1) \gt 0$ (where $\Phi$ is the standard Normal distribution function), demonstrating that it does in fact not converge.
(If you're unfamiliar with this inequality, use Calculus to minimize the function $\theta\to \Pr(|Z-\theta|\ge 1)$ (for a standard normal variable $Z$) by finding the zeros of its derivative.  You will discover the finite critical points occur where the densities at $\theta\pm 1$ are equal, immediately giving $\theta=0.$)
