Proposition 7 ("Raabe's test")

Let $(u_n)$ be a series with terms $> 0$ from some point on. If, from a certain stage, $u_{n+1}/u_n \leqslant 1 - \frac{\alpha}{n}$ for some $\alpha > 1$, then the series $(u_n)$ converges; if, from a certain point on, $u_{n+1}/u_n \geqslant 1 - \frac{1}{n}$, then the series $(u_n)$ has infinite sum.

Indeed, if $u_{n+1}/u_n \leqslant 1 - \frac{\alpha}{n}$ with $\alpha > 1$, for all $n \geqslant n_0$, then one has $u_n \preccurlyeq p_n = \prod_{k=n_0}^{n} \left( 1 - \frac{\alpha}{k} \right)$. Now, $\log \left( 1 - \frac{\alpha}{n} \right) = -\frac{\alpha}{n} - \frac{\alpha^2}{2n^2} + o \left( \frac{1}{n^2} \right)$, whence $\log p_n =$ $-\alpha \log n + k + o(1/n)$ ($k$ constant), and $p_n \sim e^k \frac{1}{n^\alpha}$; since $\alpha > 1$ the logarithmic criterion of order 0 allows one to finish.

If, on the other hand, $u_{n+1}/u_n \geq 1 - \frac{1}{n}$ from a certain point, then the same calculation proves that $u_n \geq \frac{1}{n}$, whence the proposition.

One can prove in the same way, using the logarithmic criteria of order $> 0$, the following criterion of the second kind: