Consider
\begin{align} \mathbb{P} \left( ||Y_n||_2 \leq \frac{\epsilon}{2}\right) \tag1 \end{align} where $Y_n \in \mathbb{R}^d$ is Gaussian with mean $\mu \in \mathbb{R}^d$ and covariance matrix $\frac{1}{n} I_d$, where $I_d$ is $d \times d $ identity matrix.
Assume $||\mu||_2 = \epsilon$ so that the probability in $(1)$ goes to $0$ as $n \to \infty$.
How should I accurately upper bound $(1)$ such that the bound decays to zero as $n \to \infty$?
I tried Markov Inequality as follows but it does not work.
\begin{align} \mathbb{P} \left( ||Y||_2 \leq \frac{\epsilon}{2}\right) &=\mathbb{P} \left( ||Y||_2^2 \leq \frac{\epsilon^2}{4}\right) \\ &\leq \frac{4}{\epsilon^2} \left( \mathbb{E} [||Y||_2^2] \right)\\ &= \frac{4}{\epsilon^2} \left( \sum_{l=1}^d \mathbb{E}[Y_l^2] \right)\\ &= \frac{4}{\epsilon^2} \left( \sum_{l=1}^d \left(\frac{1}{n} + \mu_l^2 \right) \right)\\ &= \frac{4}{\epsilon^2} \left( \frac{d}{n} + ||\mu||_2^2 \right) \to 4 \text{ as } n \to \infty \end{align}