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My problem: Suppose $x_n > 0$ and the infinite series $\sum x_n y_n$ converges for all nonnegative sequences $\{ y_n\}$ such that $\sum {y_n}^3$ converges. Show that $\sum x_n^{3/2}$ is convergent.

Since $\frac{1}{3} + \frac{1}{3/2} = 1$, I thought of the Holder inequality: $$\sum_{n=1}^\infty x_n y_n \leq \left(\sum_{n=1}^\infty x_n^{3/2} \right)^{2/3}\left(\sum_{n=1}^\infty y_n^{3}\right)^{1/3}$$ but the direction of the inequality is not going to help with a comparison test for convergence of $\sum x_n^{3/2}$.

WoodWorker
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2 Answers2

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Obtain a contradiction by assuming $S_n = \sum_{k=1}^n x_k^{3/2} \to \infty$ and taking $y_n = \frac{x_n^{1/2}}{S_n}$.

Note that $ \sum_{n=1}^\infty y_n^3 < \infty ,$ since

$$\sum_{n=2}^m \frac{x_n^{3/2}}{S_n^3} = \sum_{n=2}^m \frac{S_n - S_{n-1}}{S_n^3}\leqslant \sum_{n=2}^m \int_{S_{n-1}}^{S_n} \frac{dx}{x^3} = \frac{1}{2S_1^2} - \frac{1}{2S_m^2} \underset{m \to \infty}\longrightarrow \frac{1}{2S_1^2}$$

However, contrary to the hypothesis, $$\sum_{n=1}^\infty x_n y_n = \sum_{n=1}^\infty x_n \frac{x_n^{1/2}}{S_n} = \sum_{n=1}^\infty \frac{x_n^{3/2}}{S_n} = \infty,$$

using a well-known result that divergence of a positive seres $\sum_{n \geqslant 1} a_n$ implies divergence of $\sum_{n \geqslant 1} \frac{a_n}{\sum_{k=1}^n a_k}$.

RRL
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  • It's perhaps worth noting that $$\sum_{n=1}^N\frac{a_n}{\sum_{k=1}^n a_k} \geq \log\left(1+\sum_{n=1}^Na_n \right)$$ (which can be proven inductively or by recognizing the sum as an upper bound to an integral of $1/t$). I mention this since I've never heard of the result you use, but it's not hard to prove. – Milo Brandt Jan 09 '20 at 21:32
  • @MiloBrandt: Thank you. By well-known I mean it has appeared a number of times on this site. I will look for a link. – RRL Jan 09 '20 at 21:43
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    This appears to be one of the oldest references to that result. – RRL Jan 09 '20 at 21:53
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Hint: the $l^p$ norm of a sequence $x$ is equal to the sup of the $l^1$ norm of $x.y$ as $y$ ranges over $l^q$ sequences with $l^q$ norm 1.

Ciarán Ó Raghaillaigh
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