Abel–Dini theorem

Remarks

For $\alpha = 1$ and $m > n$, the terms with index from $n+1$ to $m$ satisfy $\sum_{k=n+1}^{m} a_k / s_k \ge (s_m - s_n)/s_m$. Since $s_m \to \infty$, this tends to $1$ as $m \to \infty$, so the Cauchy criterion fails. For $\alpha < 1$, one has $s_n \ge 1$ for large $n$, so $a_n / s_n^{\alpha} \ge a_n / s_n$. For $\alpha > 1$, one has $a_n / s_n^{\alpha} \le \int_{s_{n-1}}^{s_n} t^{-\alpha} \, dt$ for $n \ge 2$, and these integrals add up to at most $\int_{s_1}^{\infty} t^{-\alpha} \, dt < \infty$. Example: with $a_n = 1$, one has $s_n = n$, and the theorem says that $\sum 1/n^{\alpha}$ converges exactly when $\alpha > 1$.

Going deeper

For a student

The theorem shows that there is no slowest divergent series: from any divergent series of positive terms, $\sum a_n / s_n$ diverges, and its terms are much smaller than $a_n$. It also gives a supply of convergent series near the boundary. So no single comparison series can decide all series of positive terms, which is one reason why there is a whole scale of convergence tests.