Let $p : \mathbb{N} \to [0, 1]$ be a probability distribution over the naturals.
The Shannon Entropy is:
$$H = -\sum_{n=0}^\infty p(n)\log_2 p(n)$$
Does this series always converge?
I tried a little to attack this problem but it's been some time since I evaluate series. My attempt:
We known that $\sum_{n=0}^\infty p(n) = 1$, therefore it must be the case that $\lim\limits_{n\to\infty} p(n) = 0$, and since $\lim\limits_{x\to 0} x\log_2 x = 0$, for any $\varepsilon > 0$ there exists only a finite amount of terms greater than $\varepsilon$. The problem is that I can't conclude anything from this, because there are several series in which the terms go to zero but it still diverges.
This series does not always converge. First of all, note that the following series converges: $$\sum\limits_{i = 1}^\infty \lceil 2^i/i^2\rceil \cdot 2^{-i}.$$ This is because $\lceil 2^i/i^2\rceil \le 2^i/i^2 + 1 = O(2^{i}/i^2)$ as $i\to \infty$. Hence there exists a constant $c > 0$ such that $$\sum\limits_{i = 1}^\infty \lceil 2^i/i^2\rceil \cdot c2^{-i} = 1.$$
So, consider a probablity distribution over $\mathbb{N}$ in which for every $i\ge 1$ there are exactly $\lceil 2^i/i^2\rceil$ probablities which are equal $c2^{-i}$. Shannon entropy diverges for such distribution: \begin{align*} H = \sum\limits_{i\ge 1} \lceil 2^i/i^2 \rceil \cdot c 2^{-i} \log_2(2^i/c) &\ge \sum\limits_{i\ge \max\{1,\, 2\log_2(c)\}} \lceil 2^i/i^2 \rceil \cdot c 2^{-i} \log_2(2^i/c) \\ &\ge \sum\limits_{i\ge \max\{1,\, 2\log_2(c)\}} 2^i/i^2 \cdot c 2^{-i} \log_2(2^i/2^{i/2})\\ &= \sum\limits_{i\ge \max\{1,\, 2\log_2(c)\}} \frac{c}{2i}. \end{align*}