If an infinite sum of positive terms $\sum x_n$ diverges, must there exist a sequence $y_n$ with $y_n/x_n \to 0$ and $\sum y_n$ divergent?

Question

Suppose that $\sum_{n=1}^{\infty} x_n$ diverges, where $x_n>0$ $\forall n$. Must there exist a real sequence $y_n$ such that $y_n/x_n \to 0$ as $n \to \infty$ and such that $\sum_{n=1}^{\infty} y_n$ also diverges?

I've been stuck on this for quite a while, and have tried considering cases such as $x_n=1/n$ (for which this is possible e.g. with $y_n=1/n \log n$) to get an idea of in what situations this holds, but it doesn't seem very straightforward to generalize. I don't really know where to start in terms of approaching the problem abstractly.

Have a look at https://math.stackexchange.com/q/452053/42969. — Martin R, Jan 29 '21 at 21:20

score 4 · Accepted Answer · answered Jan 29 '21 at 22:04

Since $\sum_{k=1}^{\infty} x_{k}$ is not finite, it is possible to find blocks of integers, say $[n_{i},n_{l+1}]$ for which $\sum_{k=n_{l}}^{n_{l+1} -1} x_k >= 1$ and such that $n_{l+1} > n_{l}$.
For $k\in[n_{l},n_{l+1} -1]$ define $y_{k} = \frac{1}{l}*x_{k}$. Then $$\sum_{k=1}^{\infty} y_{k} = \sum_{l}\sum_{k = n_{l}}^{n_{l+1}-1} y_{k} >= \sum_{l}\frac{1}{l}\sum_{k=n_{l}}^{n_{l+1}-1}x_{k} >= \sum_{l}\frac{1}{l}$$

If an infinite sum of positive terms $\sum x_n$ diverges, must there exist a sequence $y_n$ with $y_n/x_n \to 0$ and $\sum y_n$ divergent?

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