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Let $(a_n)$ and $(b_n)$ be complex sequences. I don't think the following is true in general:

If $\sum a_n b_n$ converges, then $\sum \overline{a_n} b_n$ converges too.

Can someone provide a simple example to show that above statement is obviously false?

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    $\begingroup$ can you think of two sequences where each term has the same modulus, but one converges and the other doesn't? $\endgroup$
    – Zoe Allen
    Commented May 5, 2023 at 1:17
  • $\begingroup$ @ZoeAllen Thank you, I can see clearly now. $\endgroup$
    – Dilemian
    Commented May 5, 2023 at 5:42

2 Answers 2

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Let $$a_n = b_n = \frac{i^n}{\sqrt n}$$, $$a_nb_n = \frac{(-1)^n}{n}$$ and $$\overline{a_n} b_n = \frac{1}{n}$$

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Let $a_{2n} = b_{2n} = \frac{1+i}{\sqrt{4n}}$, $a_{2n+1}=b_{2n+1} = \frac{1-i}{\sqrt{4n+2}}$. Then $a_nb_n = a_n^2 = (-1)^n\frac{i}{n}$, so the series $\sum a_nb_n = i\sum(-1)^n/n$ converges by alternating series test, and $\sum\overline{a_n}b_n = \sum|a_n|^2 = \sum 1/n$ diverges.

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  • $\begingroup$ You need to add a $\sqrt{}$. $\endgroup$
    – Kroki
    Commented May 5, 2023 at 1:39
  • $\begingroup$ @Youem Edited. Thanks. $\endgroup$ Commented May 5, 2023 at 1:40

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