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If $\sum a_n$ converges, then $\sum \frac{1}{a_n}$ diverges to $\infty$

If $\sum a_n$ diverges, then for every $M > 0$ $\exists$ $n$ such that $s_n > M$

^ Are the statements above true? I'm not given any information about $a_n$, which is a series. $s_n$ denotes a partial sum of the series. I have no idea about the second one, but for the first one I'm pretty sure you can assume that any series that converges will not have its reciprocal converge, for example $\frac{1}{x^2}$ converges, but $x^2$ does not. The second one seems to be a question about whether there's a limit to the sequence, and again I have no idea how to do.

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    $\begingroup$ Neither statement is true for a general sequence $a_n$. $\endgroup$
    – Clayton
    Commented Mar 10, 2021 at 1:03
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    $\begingroup$ If $\sum a_n$ converges, then $a_n \to 0$, so $1/a_n \not\to 0$. So $\sum 1/a_n$ diverges. This is not to say $\sum 1/a_n$ diverges to $\infty$, however. $\endgroup$
    – GEdgar
    Commented Mar 10, 2021 at 1:14
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    $\begingroup$ Are you supposed to assume $a_n \ge 0$? $\endgroup$
    – Robert Israel
    Commented Mar 10, 2021 at 2:23
  • $\begingroup$ @RobertIsrael there are no assumptions to $a_n$. Nothing is known about them. $\endgroup$
    – Philip L
    Commented Mar 10, 2021 at 2:43
  • $\begingroup$ If someone can give a short answer (just like a 2 sentence explanation) then I will accept it $\endgroup$
    – Philip L
    Commented Mar 10, 2021 at 2:43

1 Answer 1

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Counterexample for (1): $a_n = -1/2^n$. $\sum_n 1/a_n$ diverges to $-\infty$, not $+\infty$, and $s_n < 0$.

Counterexample for (2): $a_n = -1$. All $s_n \le 0$.

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