convergence
Say $X_k$ is a non-negative sequence and it is known that it convergences in mean to zero. It feels like it should also convergence almost surely due to the fact that the only value a non-negative random variable can take and still average to zero is the value zero.
However, this explanation is kind of hand-wavy. Can this be proven more rigorously or disproven?
This figure shows the distributions of the means of $n=1$ (blue), $10$ (red), and $100$ (gold) independent and identically distributed (iid) normal distributions (of unit variance and mean $\mu$):
As $n$ increases, the distribution of the mean becomes more "focused" on $\mu$. (The sense of "focusing" is easily quantified: given any fixed open interval $(a,b)$ surrounding $\mu$, the amount of the distribution within $[a,b]$ increases with $n$ and has a limiting value of $1$.)
However, when we standardize these distributions, we rescale each of them to have a mean of $0$ and a unit variance: they are all the same then. This is how we see that although the PDFs of the means themselves are spiking upwards and focusing around $\mu$, nevertheless every one of these distributions is still has a Normal shape, even though they differ individually.
The Central Limit Theorem says that when you start with any distribution--not just a normal distribution--that has a finite variance, and play the same game with means of $n$ iid values as $n$ increases, you see the same thing: the mean distributions focus around the original mean (the Weak Law of Large Numbers), but the standardized mean distributions converge to a standard Normal distribution (the Central Limit Theorem).
Convergence almost surely and convergence in distribution are not the same thing.
Take a simple example where you have i.i.d. mean zero random variables $Y_i$, $i= 1, 2, \cdots$ for which the law of iterated logarithm holds. Then
$$ \lim \sup_{n \rightarrow \infty} \frac{\frac{1}{\sqrt{n}}\sum_1^n Y_i}{\sqrt{\log\log n}} = \sqrt{2}, \; a.s. $$
In particular, $\frac{1}{\sqrt{n}}\sum_1^n Y_i$ diverges almost surely, but convergence in distribution. A sequence of random variables can diverge for a fixed sample path but converge in distribution across all sample paths.
Best Answer
I thank user @guy for pointing out the mistake of my previous attempt. Instead of just deleting it, I will insert in its place a naive way to showcase what the 2nd Borel-Cantelli lemma tells us, using the case that @guy considers, a sequence of independent Bernoullis$(1/k)$.
Consider the event $\{\prod_{i=m}^k X_i = 0\}$. Since the r.v.'s are independent the probability of this event is the product of the probabilities of the individual events
$$P\left(\prod_{i=m}^k X_i = 0\right) = \prod_{i=m}^kP(X_i=0)$$
$$=\left(1-\frac 1m\right)\cdot\left(1-\frac 1{m+1}\right)\cdot...\cdot \left(1-\frac 1{k-1}\right)\cdot\left(1-\frac 1{k}\right)$$
$$=\left(\frac {m-1}m\right)\cdot\left(\frac {m}{m+1}\right)\cdot...\cdot\left(\frac {k-2}{k-1}\right)\cdot\left(\frac {k-1}{k}\right) $$
$$=\frac {m-1}{k}$$
As $k\rightarrow \infty$ this probability goes to $0$. But this probability is the probability of a subsequence to take on only the value zero. So we just concluded that the probability of an infinite subsequence of zeros is zero no matter where we start this subsequence (i.e. no matter what or how large the value of $m$ is). So the sequence does not converge to zero almost surely, even though it converges in mean to zero (and hence also in probability).
What about some intuition here? I would try this: while the single probability $P(X_k=1)$ tends to zero (and so the expected value of $X_k$ goes to zero too), it does not go "fast enough". So when looking at a whole sequence of $X_k$'s we cannot say that the sequence will converge to zero with probability one.