# Proof of convergence for the case $0<p<1$

Suppose $$(a_n)_1^\infty$$ is an increasing sequence of positive real numbers with limit $$+\infty$$. If $$p > 0$$, show that $$\sum_{n=1}^{+\infty} \frac {a_{n+1} - a_n} {a_{n+1} a_n^p}$$ converges.

Easy to do the cases $$p \geqslant 1$$: $$\sum_{n=1}^{+\infty} \frac {a_{n+1} - a_n} {a_{n+1} a_n^p} \leqslant \sum_{n=1}^{+\infty} \frac {a_{n+1} - a_n} {a_{n+1} a_n} = \frac 1{a_1} < +\infty.$$

What about the case $$0 < p < 1$$? I'm pretty sure this would not be of much trouble, but I just cannot figure it out at this moment. Any hints are welcome. Thanks in advance.

For $$0 < p < 1$$ we have, for some $$\vartheta \in (0,1)$$, $$a_{n+1}^p - a_n^p = p((1-\vartheta)a_n + \vartheta a_{n+1})^{p-1}(a_{n+1} - a_n) \geqslant p\frac{a_{n+1} - a_{n}}{a_{n+1}^{1-p}}$$ by the mean value theorem. Thus $$\frac{a_{n+1} - a_n}{a_{n+1}a_n^p} \leqslant \frac{1}{p}\cdot \frac{a_{n+1}^p - a_n^p}{a_{n+1}^pa_n^p} = \frac{1}{p}\biggl(\frac{1}{a_n^p} - \frac{1}{a_{n+1}^p}\biggr)\,.$$