0
$\begingroup$

Let $\{x_n\}$ be a given series such that it satisfies the following conditions for all sequence $\{y_n\}$ in real numbers converging to $0$. It is given that the sequence $\{y_n\}$ converges to $0$ and the series $\sum_{n=0}^{\infty} x_ny_n$ is convergent. Then show that the series $\sum_{n=0}^{\infty} |x_n|$ is convergent$,$ that is the series $\sum_{n=0}^{\infty} x_n$ is absolutely convergent.

My attempt: Since $\{y_n\}$ is converges to $0$ then for each $ε>0$ there exist a natural number $k_1$ such that $|y_n|< ε$ for all $n>k_1$. Since the series $\sum_{n=0}^{\infty} x_ny_n$ is convergent then the tail of the series goes to zero that is $x_ny_n$ tends to $0$ as $n$ tends to $\infty$. But can not proceed further to complete the proof. Please help me to solve this. Thanks in advance.

$\endgroup$

marked as duplicate by Martin R, Namaste, Did sequences-and-series Mar 15 '18 at 18:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ Not following. Suppose $y_n=0$ for all $n$. $x_n=1$ Then $y_n\to 0$ and $\sum x_ny_n$ converges, but $\sum |x_n|$ does not. $\endgroup$ – lulu Mar 15 '18 at 11:44
  • $\begingroup$ Can you clarify or edit your question? It's simply wrong as stated. Perhaps you gave the wrong assumptions? $\endgroup$ – lulu Mar 15 '18 at 11:53
  • $\begingroup$ Voting to close the question as it is wrong as stated. If you can, please edit. $\endgroup$ – lulu Mar 15 '18 at 12:12
  • $\begingroup$ Sorry at first I can't understand the question. Now I have edited it. @lulu thank you for concern about it $\endgroup$ – abcdmath Mar 15 '18 at 12:17
  • 1
    $\begingroup$ I don't understand your line of attack, though. I think the idea has to be: assume $\sum |x_n|$ diverges and construct a sequence $\{y_n\}$ which goes to $0$ for which $\sum x_ny_n$ diverges. $\endgroup$ – lulu Mar 15 '18 at 12:28
1
$\begingroup$

Suppose that $\sum |x_n|=\infty$. We will construct a sequence $\{y_n\}$ such that $y_n\to 0$ and $\sum x_ny_n$ diverges.

Since $\sum |x_n|=\infty$ we can find a strictly increasing sequence of natural numbers $N_i$ such that the partial sums satisfy $$\sum_{n=1}^{N_k}|x_n|>k$$

Now define $y_n$ by $$N_k≤n<N_{k+1}\implies y_n=\text {sign} (x_n)\times \frac 1{\sqrt{k}}$$

Clearly $y_n\to 0$. We remark that the partial sum $$\sum_{n=1}^{N_k}x_ny_n≥ \frac 1{\sqrt{k}}\sum_{n=1}^{N_k} |x_n|≥\sqrt { k}$$

Hence the partial sums for $\{x_ny_n\}$ tend to $\infty$ and we are done.

$\endgroup$
  • $\begingroup$ Thank you sir. Great one. $\endgroup$ – abcdmath Mar 15 '18 at 12:47
3
$\begingroup$

Look at $x_n=y_n = \frac{1}{n}$. Then $\sum_{n=0}^{\infty} x_ny_n$ converges, but $\sum_{n=0}^{\infty} |x_n|=\sum_{n=0}^{\infty} x_n$ is divergent !

$\endgroup$
  • $\begingroup$ Sorry for the inconvenience, but I have edited my question sir. $\endgroup$ – abcdmath Mar 15 '18 at 12:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.