Consider the usual coupon collector's problem - $n$ coupons with equal probability, and let $T_n$ be the time required to collect all $n$ coupons. It is well known that:
$$ \mathbb{E}[T_n] = nH_n $$ with $H_n$ denoting the $n$th harmonic number: $$ H_n = \sum_{k=1}^n \frac{1}{k} $$ It is easy to show that: $$ \frac{T_n}{\mathbb{E}[T_n]} \to 1 \text{ in probability} $$ This follows from applying Chebyshev's inequality and the divergence of the harmonic series: $$ \mathbb{P}\left(\left|\frac{T_n}{\mathbb{E_n}} -1 \right| > \epsilon\right) \leq \frac{\mathrm{Var}(T_n)}{n^2 (H_n)^2\epsilon^2} = \frac{n\sum_{k=1}^n\frac{n-k}{k^2}}{n^2 (H_n)^2\epsilon^2} \leq \frac{n^2(\pi^2/6)}{n^2(H_n)^2\epsilon^2} = \frac{C}{(H_n)^2}\to 0 $$ Question: Can this convergence in probability be upgraded to almost sure convergence? To see this, I tried to show that: $$ \mathbb{P}\left(\left|\frac{T_n}{\mathbb{E}[T_n]} -1 \right| > \epsilon\right) $$ is summable, which implies almost-sure convergence by Borel Cantelli. However, I am pretty sure this approach is doomed, because the estimate on the RHS will pretty much never be summable, even if one tries something less crude.
If it does not converge almost-surely, is there an elementary way to prove it does not?
It's a strange question, because there doesn't seem to be anything which indicates the joint distribution of the $T_n$ as $n$ varies, or actually to suggest that they are defined on the same probability space. In the absence of that, it's not clear what meaning to give to almost sure convergence.
If you're asking about whether you'd have almost sure convergence for such a sequence $T_n$ with independent terms, or whether you'd have almost sure convergence for every possible coupling of the distributions of the $T_n$, then (from both Borel-Cantelli Lemmas) either of these properties is in fact equivalent to the summability property you mention, namely that the probabilities $\mathbb{P}\left(\left|\frac{T_n}{\mathbb{E}[T_n]} -1 \right| > \epsilon\right)$ are summable over $n$ for every $\epsilon$.
On the other hand, since you have convergence in distribution, there must indeed exist some coupling of the random variables for which you have the desired almost-sure convergence (this follows from Skorohod's Representation Theorem).