Let $A>0$ be fixed and consider $X\_1,\ldots$ i.i.d. nonnegative random variables such that $E[1/X\_1]<\infty$.


Is is true that $$\sup\_{a\in \big (0,\frac A{\sqrt n} \big]} \sum\_{i=1}^n 1\_{X\_i>a} \frac{a^3}{X\_i^2}$$
converges in probability to $0$ ?


With the crude bound $\sum\_{i=1}^n 1\_{X\_i>a} \frac{a^3}{X\_i^2} \leq \frac {A^2}n \sum\_{i=1}^n \frac{1}{X\_i}$, the supremum is clearly $O\_P(1)$. In my research I need it to be $o\_P(1)$, but I haven't been able to prove it.



Convergence in probability of a supremum
As suggested by Anthony Quas, the supremum in question can be rewritten as
$$s\_n:=\sup\_{b\ge\sqrt n/A}S\_n(b),$$
where
$$S\_n(b):=\frac1{b^3}\sum\_{i=1}^n Z\_i^2\,1(Z\_i<b)$$
and $Z\_i:=1/X\_i$, so that the $Z\_i$'s are iid positive random variables with $EZ\_i<\infty$.


Take now any real $c>0$. Then for all large enough $n$
$$S\_n(b)=T\_n(b,c)+U\_n(b,c),$$
where
$$T\_n(b,c):=\frac1{b^3}\sum\_{i=1}^n Z\_i^2\,1(Z\_i<c),$$
$$U\_n(b,c):=\frac1{b^3}\sum\_{i=1}^n Z\_i^2\,1(c\le Z\_i<b).$$
Next, for all large enough $n$, uniformly in $b\ge\sqrt n/A$,
$$0\le T\_n(b,c)\le\frac{nc}{b^3}\frac1n\sum\_{i=1}^n Z\_i
\sim\frac{nc}{b^3}\,EZ\_1\to0$$
and
$$0\le U\_n(b,c)\le\frac n{b^2}\frac1n\sum\_{i=1}^n Z\_i\,1(c\le Z\_i)\sim\frac{n}{b^2}\,EZ\_1\,1(c\le Z\_1)
=O(EZ\_1\,1(c\le Z\_1))\underset{c\to\infty}\longrightarrow0$$
almost surely (a.s.), by the strong law of large numbers.


So, $s\_n\to0$ a.s. and hence, indeed, in probability.


