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 $\sum x_n^{3/2}$.
Best Answer
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 $\sum_{n \geqslant 1} \frac{a_n}{\sum_{k=1}^n a_k}$.