Find a sequence $y_k$ such that $\sum x_ky_k$ diverges

Question

If $\sum_{k=0}^\infty |x_k|=\infty$, could we find a sequence $y_k\rightarrow 0$ such that $\sum x_ky_k$ diverges?

Answer 1

Assume that each $x_k \ge 0$. Otherwise just add $\pm$ to the $y_k$ as necessary.

Choose $n_1$ so that $\displaystyle \sum_{k=1}^{n_1} x_k \ge 1$. Define $$y_1 = y_2 = \cdots = y_{n_1} = 1.$$ Choose $n_2$ so that $\displaystyle \sum_{k=n_1+1}^{n_2} x_k \ge 2$. Define $$ y_{n_1+1} = y_{n_1+2} = \cdots = y_{n_2} = \frac 12.$$ Choose $n_3$ so that $\displaystyle \sum_{k=n_2+1}^{n_3} x_k \ge 3$. Define $$ y_{n_2+1} = y_{n_2+2} = \cdots = y_{n_3} = \frac 13.$$

And so on.

Answer 2

Yes.

Let $S_n:=42+\sum_{k=0}^n |x_k|$ for $n\ge 0$ and $S_{-1}:=42$, so that $|x_n|=S_n-S_{n-1}$ for all $n\ge 0$. We see that $S_n>0$ for $n\ge -1$ and are given that $S_n\to \infty$ as $n\to \infty$.

This allows us to define $$y_n=\frac{\operatorname{sgn} x_n}{\sqrt{S_n}+\sqrt{S_{n-1}}}$$ for $n\ge 0$.

Note that $S_n\to\infty$ implies $y_n\to 0$. We compute $$ x_ny_n=\frac{|x_n|}{\sqrt{S_n}+\sqrt{S_{n-1}}}=\frac{S_n-S_{n-1}}{\sqrt{S_n}+\sqrt{S_{n-1}}}=\sqrt{S_n}-\sqrt{S_{n-1}}.$$ Thus by teleskoping, $$\sum_{k=0}^nx_ky_k=\sqrt{S_n}-\sqrt{42}$$ and this diverges $\to \infty$ as $n\to\infty$.

Answer 3

A slight variation of Umberto's excellent answer but with a closed formula for $y_n$ in terms of $s_n=\sum_{k=0}^{n} |a_k|$.

We can assume that $x_0\neq 0$, because if we solve for that case, we can find some $n_0$ with $x_{n_0}\neq 0$, and solve for $x_k^{'}=x_{n_0+k}$ and then choose $y_k=x_{k-n_0}$ for $k\geq n_0$ and $y_k=0$ for $n<n_0.$

You can then use:

$$y_{n} =\dfrac{\operatorname{sgn}x_n}{\sqrt{s_n}}\tag{1}$$

(For the purposes of this question, we'll define $\operatorname{sgn} 0 = 1.)$

These $y_n$ suffice because $x_ny_n\geq 0$ for all $n$, and the $|y_n|$ are decreasing, so we get $$\sum_{k=0}^{n} x_ky_k=\sum_{k=0}^n |x_k||y_k|\geq |y_n| s_n=\frac{s_n}{\sqrt{s_n}}=\sqrt{s_n}$$

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We can technically define $y_n=0$ when $x_n=0$, but then we need to be careful about why $\sum_{k=0}^{n} x_ny_n\geq \sqrt{s_n}.$ When $x_{n}=0$, we'd have (by an induction argument:) $$\sum_{k=0}^{n}x_ky_k =\sum_{k=0}^{n-1}x_ky_k \geq \sqrt{s_{n-1}}=\sqrt{s_{n}}$$

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Instead of $\sqrt{x}$, you can choose any positive $f(x)$ defined on $(0,+\infty)$ which is (not necessarily strictly) increasing and $\lim_{x\to\infty} f(x)=\infty$ and $\lim_{x\to\infty} \frac{x}{f(x)}=\infty$ and then define:

$$y_n=\dfrac{\operatorname{sgn}x_n}{f(s_n)}$$

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The relationship to Umberto's answer was that I realized you could alter his definition so that $y_n=\frac{1}{k}$ where $k$ is the largest value so that $s_n\geq \frac{k(k-1)}{2}$.

This in turn gives

$$y_n=\dfrac{1}{\left\lfloor \dfrac{\sqrt{8s_n+1}+1}2\right\rfloor}$$

This is not the same value as Umberto's, but it is related. It is slightly less, so I wondered if this sufficed.

If it sufficed, though, it meant that we could also use $y_n'=\frac{1}{\sqrt{s_n}}$ as the values, since it $y_n'\Theta(y_n).$ So I checked the simpler $\frac{1}{\sqrt{s_n}}$ case and got this result.