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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.

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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 $$\frac{a_{n+1} - a_n}{a_{n+1}a_n^p} \leqslant \frac{1}{p}\cdot \frac{a_{n+1}^p - a_n^p}{a_{n+1}^pa_n^p} = \frac{1}{p}\biggl(\frac{1}{a_n^p} - \frac{1}{a_{n+1}^p}\biggr)\,.$$

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  • $\begingroup$ Oh right, thanks so much!👍👍👍👍 $\endgroup$
    – xbh
    Commented Dec 4, 2019 at 13:39

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