convergence divergence - Show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant $2p$

Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = \frac{1}{n}\sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{n\geq 1}$ converges in probability to the constant 2p.

I found that $E[T_n] = 2p$ and that $\operatorname{Var}[T_n] = 2p(1-p)\frac{2n-1}{n^2}$.

My definition of convergence in probability is the following:

$$\forall \epsilon > 0 \space\mathbb{P}(\vert T_n - 2p \vert > \epsilon) \to 0$$

I can also use the following criterion:

Convergence in probability iff

$$\lim_{n\to\infty} \mathbb{E}\Big[\frac{\vert T_n - 2p\vert}{\vert T_n - 2p\vert + 1}\Big] = 0$$

To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?

Answer

Claim. If $\mu_n = \mathbf{E}(T_n) \to \mu$ and $\sigma_n^2 = \mathbf{V}\mathrm{ar}(T_n) \to 0$ then $T_n \to \mu$ in $\mathscr{L}^2$ and, hence, in probability too.

Proof. We have $\mathbf{E}(|T_n - \mu|^2) = \mathbf{E}(|T_n - \mu_n|^2) + 2(\mu_n - \mu) \mathbf{E}(T_n - \mu_n) + (\mu_n - \mu)^2 \to 0.$ Q.E.D.