If I have $x_1, ..., x_k=o(n)$ and $j=O(1)$.
Is it possible to prove something like:
$$\sum_{i=1}^k {n \choose j} \left(\frac{x_i}{n}\right)^j \left(1-\frac{x_i}{n}\right)^{k-j} \sim {n \choose j} \left(\frac{x}{n}\right)^j \left(1-\frac{x}{n}\right)^{k-j}$$
where $x:=\sum_{i=1}^k x_i$.
Is there any hope to prove that or should I let it be?
Or could one approximate this term by some other Definition for $x$?
The simplified term is then: $$\sum_{i=1}^k x_i^j \left(n-x_i\right)^{k-j} \sim x^j \left(n-x \right)^{k-j}$$
By exploiting the convexity/concavity (depending on $j$) of $(x(1-x))^j$ over $[0,1]$, you always have an inequality in a certain direction through the Jensen's inequality, but to prove that your inequality holds even in the opposite direction (up to a constant factor) you need stronger assumptions con the distribution of the $x_i$s.