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If $(a_n), (b_n)$ are sequences then

$1.$ If $\lim_{n\rightarrow\infty} a_n = \infty$ and $\lim_{n\rightarrow\infty} b_n = L>0$ then $\lim_{n\rightarrow\infty} a_nb_n = \infty$. True

$2.$ If $\lim_{n\rightarrow\infty} a_n = \infty$ and $\lim_{n\rightarrow\infty} b_n = -\infty$ then $\lim_{n\rightarrow\infty} (a_n+b_n) = \infty$. False since $\infty-\infty$ is undefined.

$3.$ If $\lim_{n\rightarrow\infty} a_n = \infty$ and $\lim_{n\rightarrow\infty} b_n = -\infty$ then $\lim_{n\rightarrow\infty} a_nb_n = -\infty$. True

$4.$ If neither $(a_n), (b_n)$ converge then $(a_nb_n)$ doesn't converge. False since $(\frac{1}{n})$ and $n$ doesn't converge individually but when multplied they converge to 1.

$5.$ If $(|a_n|)$ converge then $(a_n)$ converge. False since $(|(-1)^n|)$ converge but $((-1)^n)$ doesnt converge. However the converse of this statement is true.

$6.$ If $a_n=0$ for every $n$ then $\sum a_n$ converges. True. Sum converges to $0$.

$7.$ If $\sum a_n$ converges then $\sum \frac{1}{a_n}$ diverges to infinity. True

$8.$ If $\sum a_n$, $\sum b_n$ converges then $\sum (a_n + b_n)$ converges. True

$9.$ If $a_n>c>0$ for every $n$ then $\sum a_n$ diverges to infinity. True

$10.$ If $\sum a_n$ diverges and $(b_n)$ is bounded then $\sum a_nb_n$ diverges to infinity. True

$11.$ If $a_n>0$ and $\sum a_n$ converges then $\sum (a_n)^2$ converges. True.

Are these answers correct?

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  • $\begingroup$ What question are you asking? $\endgroup$ Commented Jul 19, 2017 at 16:31
  • $\begingroup$ Edited the title. $\endgroup$ Commented Jul 19, 2017 at 16:32
  • $\begingroup$ But you are supposed to ask only one question at a time. Additionally, only six questions can be asked per day. $\endgroup$ Commented Jul 19, 2017 at 16:33
  • $\begingroup$ So I am supposed to make 11 different posts for just a confirmation of true and false. Besides my only question is whether these answers are correct or not. $\endgroup$ Commented Jul 19, 2017 at 16:34
  • $\begingroup$ _Yes..._${}{}{}$ $\endgroup$ Commented Jul 19, 2017 at 16:35

2 Answers 2

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Most of your answers look alright. A few remarks:

$7.$ If $\sum a_n$ converges then $\sum \frac{1}{a_n}$ diverges to infinity. True

Do you know anything about the sign of $a_n$? If not, note that the convergence of $\sum a_n$ implies that $a_n \to 0$, but not that $a_n > 0$; i.e. it's possible that $a_n <0$.

Also note that it doesn't even have to diverge to $\pm\infty$, for example: $$a_n = \frac{(-1)^n}{n} \implies \frac{1}{a_n} = (-1)^nn$$

$10.$ If $\sum a_n$ diverges and $(b_n)$ is bounded then $\sum a_nb_n$ diverges to infinity. True

If $\sum a_n$ diverges, it doesn't necessarily diverge to infinity (think of $1-1+1-1+1-\ldots$ for example); now take $b_n \equiv 1$.

And as pointed out by Akiva Weinberger in the comment, you can even make it converge (e.g. take $b_n = \tfrac{1}{n}$ and keep $a_n$ as above).

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  • $\begingroup$ So for 7 the correction is that divergence can be to positive or negative infinity? $\endgroup$ Commented Jul 19, 2017 at 16:40
  • $\begingroup$ For #10, you can even make $\sum a_nb_n$ converge. $\endgroup$ Commented Jul 19, 2017 at 16:40
  • $\begingroup$ for number 7, suppose $a_n = (-\frac 12)^n$ then $\sum \frac 1{a_n} = \sum (-2)^n $ wich clearly diverges but not to infinity or negative infinity. $\endgroup$
    – Doug M
    Commented Jul 19, 2017 at 16:44
  • $\begingroup$ #7; indeed, I was adding a different example to my answer. $\endgroup$
    – StackTD
    Commented Jul 19, 2017 at 16:45
  • $\begingroup$ @AkivaWeinberger Indeed, I added this. $\endgroup$
    – StackTD
    Commented Jul 19, 2017 at 16:48
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For 2., "$\infty-\infty$ is undefined" is not an acceptable justification. You need to actually produce two sequences $a_n$ and $b_n$ with the stated properties and then show that $\lim(a_n+b_n)$ is not infinite. Also, 10. is not correct because you can take $b_n = 0$, for instance.

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  • $\begingroup$ I would say "indeterminate" and not undefined. $\endgroup$
    – Doug M
    Commented Jul 19, 2017 at 16:41
  • $\begingroup$ @DougM That's a fair criticism. I was only quoting the OP. $\endgroup$
    – Alex Ortiz
    Commented Jul 19, 2017 at 16:44

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