If there a way to simplify the condition that $\mathbb{E}\left(\inf\{n : X_n > n\}\right) < \infty$, where $X_n$ is i.i.d.? If $X_n$ is an i.i.d. sequence of random nonnegative variables, I recently proved a lemma with the conclusion holding if and only if $$
\mathbb{E}\left(\inf\{n\in\mathbb{N} : X_n > n\}\right) < \infty
$$
This isn't the worst, but it's a bit difficult to look at and understand. I'm wondering if there's an easier equivalent condition, especially since this is a property of the distribution and strictly speaking the sequence irrelevant. Is there some way to frame this without having to construct an infinite sequence. Ideally something of the form $\mathbb{E}(f(X_1))<\infty$.
 A: This is not a complete characterisation, but if we define $\tau = \inf\{n : X_n > n\}$, then we can show that
$$
\mathbf{E} \tau
\begin{cases}
< \infty & \text{if } \liminf_{n\rightarrow\infty} n \Pr(X_1 > n) > 1, \\
= \infty &\text{if } \limsup_{n\rightarrow\infty} n \Pr(X_1 > n)< 1
\end{cases}
$$
This is very explicit in terms of the tail probabilities, but leaves out only the case $\Pr(X \geq n) \sim 1/n$, which you may or may not be satisfied with.

To see this, define.
First, we notice that
\begin{align*}
\Pr(\tau = \infty)
= \prod_{n=1}^\infty \Pr(X_n \leq n)
= \prod_{n=1}^\infty (1 - \Pr(X_1 > n)). 
\end{align*}
It is a standard fact about the convergence of infinite producs that this converges to $0$ if and only if
$$
\sum_{n=1}^\infty \Pr(X_1 > n)
= \infty,
$$
in which case $\mathbf{E} X_1 = \infty$.
Thus, if $\mathbf{E} X_1 < \infty$ (in this case, $\limsup_{n\rightarrow\infty} n \Pr(X_1 > n) = 0$), we have that $\Pr(\tau = \infty) > 0$ and so that $\mathbf{E} \tau = \infty$.
In the $\mathbf{E}X_1 < \infty$ case, we have that $\Pr(\tau = \infty) = 0$, and so
\begin{align*}
\mathbf{E} \tau
= \sum_{n=0}^\infty \Pr(\tau > n) 
= \sum_{n=0}^\infty \Pr(X_1 \leq 1, \dotsc, X_n \leq n)
= \sum_{n=0}^\infty \prod_{k=1}^n \Pr(X_1 \leq k).
\end{align*}
Now, to see if this series converges, we can use something like Raabe's test (https://mathworld.wolfram.com/RaabesTest.html), which in our case yields
\begin{align*}
\rho_n
&:= n \left(\frac{\prod_{k=1}^n \Pr(X_1 \leq k)}{\prod_{k=1}^{n+1}\Pr(X_1 \leq k)} - 1\right) \\
&= n \frac{\Pr(X_1 > n+1)}{\Pr(X_1 \leq n+1)} \\
&\sim n \Pr(X_1 > n),
\end{align*}
so that the series converges or diverges if $\liminf \rho_n > 1$ or $\limsup \rho_n < 1$ respectively.
But this is exactly the initially-stated condition.
