Let $p_k \ge 0 $ be a probability distribution over the positive integers: $\sum_{k=1}^\infty p_k = 1$
Let $ \mu_n = \sum_{k=1}^n k \, p_k$, so that $\mu_{\infty}=\mu$ is its mean - it the limit exists.
We know that $\mu_n$ does not necessarily converge, there are discrete distributions with infinite mean. For example, for $p_k = \frac1{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}$, we have $\mu_n= \log n + O(1)$
My question is : what can we say about the asymptotics of $\mu_n$ in general?
In particular, can this be proven or disproven? $$\lim_{n \to \infty }\frac{1}{n} \mu_n =0$$
Yes, this limit is $0$. Fix $\varepsilon >0$. Since $\sum_{k=1}^\infty p_k$ is convergent there is some $n_0$ such that $\sum_{k=n_0+1}^\infty p_k < \varepsilon/2$. For $n>n_0$ we have $$ 0 \le \frac{1}{n}\mu_n= \frac{1}{n}\sum_{k=1}^{n_0} kp_k + \sum_{k=n_0+1}^{n} \frac{k}{n}p_k \le \frac{1}{n}\sum_{k=1}^{n_0} kp_k + \sum_{k=n_0+1}^n p_k \le \frac{1}{n}\sum_{k=1}^{n_0} kp_k + \frac{\varepsilon}{2}. $$ Now choose $n_1 > n_0$ such that $\frac{1}{n_1}\sum_{k=1}^{n_0} kp_k < \varepsilon/2$. For $n\ge n_1$ we get: $$ 0 \le \frac{1}{n}\mu_n \le \frac{1}{n_1}\sum_{k=1}^{n_0} kp_k + \frac{\varepsilon}{2} < \varepsilon. $$