# Convergence of series/absolute convergence

Let $y_n$ be a sequence of real numbers such that for all sequences of real numbers $x_n$ with $\lim x_n =0$ the series $\displaystyle \sum_{n=1}^{\infty} x_n y_n$ converges. Prove that $\displaystyle \sum_{n=1}^{\infty} |y_n|$ converges.

I deleted my solution... All i know/extract from the exercise is that $\lim x_n y_n =0$ and $\lim x_n =0$. Now I'm stuck. I don't know how to continue. Any hints appreciated. I know many criterions/ tests that should help me examine if the series converges, yet I cannot apply them.. because I do not know if $\lim |y_n| =0$. If it does not the series clearly diverges.

• It's not true that if $x_ny_n$ and $x_n$ tend to $0$, then $y_n$ tends to 0 – Bernard Jan 3 '15 at 13:19
• Yes.. you can find counterexamples.. ! You're right.. I guess I have to reconsider there – Tolaso Jan 3 '15 at 13:20
• Now, that I take a better look at the exercise.. I don't where to start... !! Since the other observation I made was useless.. Any hints? – Tolaso Jan 3 '15 at 13:54
• You note that your proposed solution is wrong. You should probably edit your question to remove that "solution" and add any other work you have done so far on the problem. – Rory Daulton Jan 3 '15 at 13:58
• It is interesting to note that this means that $\sum{1\over\log n}y_n$ converges, along with many other sequences where $x$ approaches $0$ even more slowly. Perhaps this is an approach you could take: construct a sequence of sequences of $x_n$ each of which converge to $0$ but increasingly slowly... – abiessu Jan 3 '15 at 14:07

## 1 Answer

Here is how to see $\lim |y_n| =0$: if the sequence of non-negative terms $|y_n|$ does not converge to $0$, then there is $\epsilon > 0$ and a sequence of indices $n_1 < n_2 < \ldots$, such that $|y_{n_k}| > \epsilon$ for $k = 1, 2, \ldots$. Define a sequence $x_n$ such that

$$x_{n_k} = \frac{\mathrm{sgn}(y_{n_k})}{k}$$

and $x_n$ is $0$ if $n$ is not one of the $n_k$. Then $\lim x_n = 0$, but $\sum_{n=1}^{\infty}x_ny_n$ is not convergent, as the subsequence comprising its non-zero elements is bounded below by the harmonic sequence $\frac{\epsilon}{k}$. This contradicts our assumptions, so we must have $\lim|y_n| = 0$.

To answer the main question, assume $\sum_{n=1}^{\infty}|y_n|$ does not converge, then there is a sequence of indices $N_1< N_2 < \ldots$ such that the partial sum $\sum_{n=1}^{N_k}|y_n| > k$, $k = 1, 2 \ldots$. Now define $$x_n = \frac{\mathrm{sgn}(y_{n})}{\sqrt{k_n}}$$ where $k_n$ is the least $k$ such that $n \le N_k$. then $\lim x_n = 0$, but $\sum_{n=1}^{N_k}x_ny_n > \frac{k}{\sqrt{k}} = \sqrt{k}$, so $\sum_{n=1}^{\infty}x_ny_n$ is not convergent, contradicting our assumptions.

• Ok.. but this answers partially the problem... How to prove that thse series converges? It is not necessary that if the sequence goes to zero, then the series converges. Example is $1/n$. – Tolaso Jan 3 '15 at 15:06
• You said you needed to know that $\lim{y_n} = 0$, I answered that part of your question, because I thought you meant you could take it from there. – Rob Arthan Jan 3 '15 at 15:08
• Well, If I knew that the sequence was zero (which you showed very nicely) then I would be able to apply some tests. But , playing around with the tests well I did not get anything useful. Or for example use the following sentence: $$\lim \left| \frac{y_{n+1}}{y_n} \right| =\ell \implies \lim \sqrt[n]{y_n} =\ell.$$ – Tolaso Jan 3 '15 at 15:10
• Hint: the approach of the above proof works for your main question: assume $\sum|y_n|$ diverges and use that to design a sequence $x_n$ such that $\sum x_ny_n$ diverges. – Rob Arthan Jan 3 '15 at 15:15
• I'll try that... thank you for the moment. – Tolaso Jan 3 '15 at 15:23