The problem:
Let $P_n$ and $Q_n$ be the distribution of the mean of a sample of size $n$ from the $N(0, 1)$ and the $N(\theta_n, 1)$ distribution, respectively. Show that $P_n$ and $Q_n$ are mutually contiguous if and only if $\theta_n=O(1/\sqrt{n})$.
My attempt goes like this. In my book, it says that if $P_n$ and $Q_n$ are such that $dP_n/dQ_n$ converge in distribution to $\exp{N(\mu, \sigma^2)}$, they are mutually contiguous if and only if $\mu=-\sigma^2/2$.
$dP_n/dQ_n$ is the likelihood ratio, so $$ \frac{dP_n}{dQ_n}=\frac{\exp\left\{-\frac{1}{2}\sum_{i=1}^nX_i^2\right\}}{\exp\left\{-\frac{1}{2}\sum_{i=1}^n(X_i-\theta_i)^2\right\}}=\exp\left\{\frac{1}{2}\left(\sum_{i=1}^n\theta_i^2-2\sum_{i=1}^nX_i\theta_i\right)\right\} $$
Assume that $\theta_n=O(1/\sqrt{n})$, meaning that $\theta_n=\mu_n/\sqrt{n}$ where $\mu_n=O(1)$. Define $Y_n=-2X_n\mu_n$. Then (under $Q_n$) $$ X_n \sim N\left(\frac{\mu_n}{\sqrt{n}}, 1\right) \Longrightarrow Y_n =-2X_n\mu_n \sim N\left(\frac{-2\mu_n^2}{\sqrt{n}}, 4\mu_n^2\right) \Longrightarrow Y_n/\sqrt{n}\sim N\left(\frac{-2\mu_n^2}{n}, \frac{4\mu_n^2}{n}\right) $$ and the sum can be written as $\sum_i -2X_i\theta_i=\sum_i -2X_i\frac{\mu_i}{\sqrt{i}}=\sum_i Y_i/\sqrt{i}$.
Now, what I wanted is actually there - the mean of $Y_n/\sqrt{n}$ is minus one half of the variance. But some questions remain on how to complete this:
What can I say about the limit of this? Not sure how to approach it. I would guess that I should use the $O(1\sqrt{n})$ in some way to make sure that that assumption is necessary.
What does $O(1/\sqrt{n})$ mean for the first sum, where we have a quadratic term? Should I simply use proportionality since it's a constant?
Any hints are much appreciated!