Question from Putnam '08: Given $F_n(x)$, find $\lim_{n\to\infty}\frac{n!F_n(1)}{\ln(n)}$ 
Problem Statement: Let $F_0(x) = \ln(x)$.  For $n\ge0$ and $x\gt0$, let $F_{n+1}(x) = \int_0^xF_n(t)dt$.  Evaluate
  $$\lim_{n\to\infty}\frac{n!F_n(1)}{\ln(n)}$$
  Source: Putnam 2008, Problem B2.


My solution:
First, show by induction that:
$$F_n(x) = \frac{x^n}{n!}\left(\ln(x) - H_n\right)$$
...where $H_n$ is the $n$th harmonic number.
(Induction proof omitted because it is trivial, and doesn't relate to my question)
Then, the limit becomes:
$$\begin{align}
\lim_{n\to\infty}\frac{n!F_n(1)}{\ln(n)} &= \lim_{n\to\infty}\frac{n!}{\ln(n)}\frac{1^n}{n!}\left(\ln(1) - H_n\right) \\
&= \lim_{n\to\infty}\frac{\left(\ln(1) - H_n\right)}{\ln(n)}\\
&= \lim_{n\to\infty}\frac{-H_n}{\ln(n)}\\
&= \lim_{n\to\infty}\frac{-H_n}{\ln(n)} \tag{1}\\
&= -\lim_{n\to\infty}\frac{\ln(n) + \gamma}{\ln(n)} \tag{2}\\
&\overset{\text{L'H}}{=} -\lim_{n\to\infty}\frac{\frac{1}{n}}{\frac{1}{n}} \tag{3}\\
&= -1 \tag{4}\\
\end{align}$$
(where $\gamma$ is Euler's Constant)

My Questions


*

*For Putnam-style grading, do I have to prove the limit exists by definition (e.g. $\epsilon-\delta$ proof)?

*Do I have to show that I can apply limit rules (e.g. L'Hopital's from $(2)$ to $(3)$) to limits of discrete variables?

*In between lines $(1)$ and $(2)$, I made a jump based on the definition that $\lim_{n\to\infty}(H_n - \ln(n)) = \gamma$  So, as $n$ becomes large, $H_n$ is approximately $\ln(n) + \gamma$.  I am most uneasy about this step, as it feels like a "back of the paper" sort of substitution and not a rigorous one.  Is this a justified step?


And, of course, I'd be open to suggestions/comments on the style of my solution and general rigor, but the above three points are the main ones to which I'm looking for answers.
 A: I feel like your perception of this problem may be backwards: to my mind, the induction to prove the form of $F_n(x)$ is the 'meat' of the problem, and once you've got that result it's the rest of the problem that's trivial.  To answer your specific questions, though: you shouldn't need an epsilon-delta proof for limits on a Putnam; once you have an explicit form for $F_n$ (and note that the first integral is improper so a little justification may help there), you can manipulate the limit quite a bit — as long as you don't do anything improper, you should be fine.
In this case, if you didn't know the form of the Harmonic series explicitly (and I would personally take $H_n = \ln n+O(1)$ as well-enough established that it didn't need independent justification, but I wouldn't fault someone for feeling otherwise) then you can use Riemann estimates for $\int_1^n \frac1x dx$ to bound it: just break it up as $\sum_{i=1}^{n-1}\left(\int_i^{i+1}\frac1xdx\right)$ and note that the integral in parentheses is bounded between $\frac1{i+1}$ and $\frac1i$.  Summing, this gives $\ln n\leq H_n\leq \ln n+1$, and that's more than enough to give the result: since $F_n(1) = -\frac{H_n}{n!}$ then $-\frac{\ln n+1}{n!}\leq F_n(1)\leq -\frac{\ln n}{n!}$ and so $-\left(1+\frac1{\ln n}\right)\leq \frac{n!F_n(1)}{\ln n}\leq -1$; the squeeze here is trivial, and you don't need L'Hopital's rule at all.
In general, rigor in contest problems is to be encouraged, but it should also be the last thing you work on; for an exam like the Putnams where you (almost certainly) won't be able to complete all the problems, putting effort into a new problem is IMHO more likely to bear fruit (and points) than the last few drops of rigor on a problem you've already gotten a result for.
