Does the expression
$$\ln\prod_{i = 1}^n\left(1 + \frac{x_i^2}{n}\right)$$
have either
- a "nicer" closed-form, or
- quantitative upper/lower bounds on such a form?
For me, "nice" essentially means "simple functions of $n, \lVert \vec x\rVert_2$". As an example, one can write
$$\ln\prod_{i = 1}^n\left(1 + \frac{x_i^2}{n}\right) = \frac{\lVert \vec x\rVert_2^2}{n} + O(1/n^2)$$
I would say this is a qualitative "nice" approximation of my function (to first-order in $n$, where $O(\cdot)$ tracks dependency on $n$, but not $x_i$). I'm interested if this can be made quantitative, and in particular if the dependency on $\lVert\vec x\rVert_2$ can be tracked in the error term $O(1/n^2)$, and there can be no implicit constants.
One can prove the bounds
$$ \frac{\lVert \vec x\rVert_2^2}{n} - \frac{\lVert \vec x\rVert_2^4}{2n^2} \leq \ln\prod_j(1+\frac{x_j^2}{n}) \leq \frac{\lVert \vec x\rVert_2^2}{n}, $$ e.g. "nice" in the way I wanted. This is done by chaining together some standard inequalities. For the upper bound, write
$$ \ln\prod_j(1+\frac{x_j^2}{n}) = \frac{1}{n}\ln\prod_j(1+\frac{x_j^2}{n})^n \leq \frac{1}{n} \ln\prod_j\exp(x_j^2) = \frac{1}{n}\ln\exp(\lVert \vec x\rVert_2^2) = \frac{\lVert \vec x\rVert_2^2}{n}. $$
For the lower bound, note that
$$\prod_j(1+\frac{x_j^2}{n})\geq 1 + \frac{\lVert \vec x\rVert_2^2}{n}$$
Combined with the bound $\ln(1+x)\geq x-\frac{x^2}{2}$ for $x\geq 0$, one gets the bound that I claimed above.