Let $\mu_n$ and $\sigma_n$ be the mean and standard deviation of $X_n$. Suppose there is a sequence of positive numbers $k_n$ such that $\sum_n 1/k_n^2 < \infty$ while $\sum_n (\mu_n - k_n \sigma_n) = + \infty$. By Chebyshev's inequality, $\mathbb P(X_n \ge \mu_n - k_n \sigma_n) > 1 - 1/k_n^2$, so $\sum_n 1/k_n^2 < \infty$ implies a.s. $X_n \ge \mu_n - k_n \sigma_n$ for all sufficiently large $n$, and thus a.s. $\sum_n X_n = +\infty$.

Let $\mu_n$ and $\sigma_n$ be the mean and standard deviation of $X_n$. Suppose there is a sequence of positive numbers $k_n$ such that $\sum_n 1/k_n^2 < \infty$ while $\sum_n (\mu_n - k_n \sigma_n) = + \infty$. By Chebyshev's inequality, $\mathbb P(X_n \ge \mu_n - k_n \sigma_n) > 1 - 1/k_n^2$, so $\sum_n 1/k_n^2 < \infty$ implies a.s. $X_n \ge \mu_n - k_n \sigma_n$ for all sufficiently large $n$, and thus a.s. $\sum_n X_n = +\infty$.

Analogous conditions could be formulated in terms of other moments.