Let $\alpha \in (0,\infty)$ and $(u_n)$ a sequence of positive real numbers.

Suppose that $\sum_{n\geq 0} u_n$ converges.

Prove that $\displaystyle \sum_{n\geq 0} \frac{(u_n)^\alpha}{n}$ converges.

Cauchy Schwarz inequality yields $\displaystyle\sum_{n=0}^M\frac{(u_n)^\alpha}{n} \leq \sqrt{\sum_{n=0}^M(u_n)^{2\alpha} \sum_{n=0}^M\frac{1}{n^2}} $

  • When $\alpha \geq \frac{1}{2}$, using that $u_n \to 0$ yields the convergence of $\sum_{n=0}^\infty(u_n)^{2\alpha}$.

Thus the series $\displaystyle \sum_{n\geq 0} \frac{(u_n)^\alpha}{n}$ has bounded partial sums and $\displaystyle\frac{(u_n)^\alpha}{n} \geq 0$.

This proves that $\displaystyle \sum_{n\geq 0} \frac{(u_n)^\alpha}{n}$ converges.

  • What do when $\alpha < \frac12$ ?

A generalisation of the Cauchy-Schwarz inequality is Hölder's inequality,

$$\int \lvert f(t)g(t)\rvert\,dt \leqslant \left(\int \lvert f(t)\rvert^p\right)^{1/p} \left(\int \lvert g(t)\rvert^{p/(p-1)}\right)^{(p-1)/p}$$

if we write it for integrals. You can write it for sums or regard the sums as integrals with respect to the counting measure, whatever you prefer.

  • $\begingroup$ It works, thanks. I wonder if there's way without Hölder though. $\endgroup$ – Gabriel Romon Sep 14 '14 at 19:55
  • $\begingroup$ None obvious (to me). There are probably other ways, but I don't see one right now. $\endgroup$ – Daniel Fischer Sep 14 '14 at 20:01
  • 2
    $\begingroup$ Just Young's inequality on it's own will do the trick; for $\alpha \geq 1$ we are just making the terms smaller (eventually), so we only need to consider the case $\alpha < 1$. Take $\beta > 1$ such that $\alpha + 1/\beta = 1$, then get $$\frac{(u_n)^\alpha}{n} \leq \alpha u_n + \frac{1}{n^{\beta} \beta},$$ which completes the proof. (Though saying this, Young's inequality is logically equivalent to Hölder's inequality :P) $\endgroup$ – Matt Rigby Sep 14 '14 at 20:31

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