In Statistical Inference, we were taught this theorem,
Consider an estimator $T_n$ of population parameter $\theta$, using $n$ samples. $T_n$ is a Consistent Estimator of $\theta$ if $$E[T_n] \to \theta \text{ and $Var(T_n) \to 0$}$$
However the converse of the above is not true. To show this, I made the following estimator of $\mu$. Let $X_1, X_2...X_n$ be random samples from a population with finite mean $\mu$ and varaice $\sigma^2$. Denote the sample mean by $\bar{X_n}$.
$$T_n = \bar{X_n} + n^2Y_n$$ Where $$Y_n \sim Ber(\frac{1}{n})$$ and $Y_i$'s are independent of the sample $X_1, X_2...X_n$. It's clear $E[T_n] = \mu + n$. Clearly, $E[T_n] \not\to \mu$. Using the Weak Law of Large numbers. $$(1 - \frac{\sigma^2}{n \epsilon^2}) \leq P(|\bar{X}_n - \mu| \leq \epsilon) \leq 1$$ Noting that $Y_n$ is $0$ with probability $(1 - \frac{1}{n})$, and in that case, $T_n | (Y_n = 0) =\bar{X}_n$. We can say
$$(1 - \frac{\sigma^2}{n \epsilon^2})(1 - \frac{1}{n}) \leq P(|T_n - \mu| \leq \epsilon) \leq 1$$ Taking the limit $n \to \infty$, we get $T_n$ is a consistent estimator of $\mu$.
Can someone produce an example purely using the samples (not like the $Y_n$ I used here, as I'm not sure how we'll generate the "randomness" $Y_n$ assures)?
An easy example is to sample from a Pareto distribution with parameter $\alpha \in (1, 2]$. In this way the sample mean $\bar{X}_n$ is an unbiased and consistent estimator whose variance does not tend to $0$. Indeed: