Proof of a series $\sum_{n=0}^{\infty} x_n$ is absolutely convergent. [duplicate]

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.

marked as duplicate by Martin R, Namaste, Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 15 '18 at 18:35

• 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. – lulu Mar 15 '18 at 11:44
• Can you clarify or edit your question? It's simply wrong as stated. Perhaps you gave the wrong assumptions? – lulu Mar 15 '18 at 11:53
• Voting to close the question as it is wrong as stated. If you can, please edit. – lulu Mar 15 '18 at 12:12
• Sorry at first I can't understand the question. Now I have edited it. @lulu thank you for concern about it – abcdmath Mar 15 '18 at 12:17
• 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. – lulu Mar 15 '18 at 12:28

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.

• Thank you sir. Great one. – abcdmath Mar 15 '18 at 12:47

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 !

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