I was told that due to sequence limit arithmetic, suppose ${a_n}$ and $ {a_n + b_n}$ converges, therefore ${b_n}$ converges since:

$$ \lim_{n\to\infty}({a_n}+{b_n}) = \lim_{n\to\infty} a_n + \lim_{n\to\infty} b_n $$

from here: Is this proof correct? proving a sum of a convergent and divergent sequences is a divergent sequence

But then I came to this exercise which contradicts what I was told:

1.Prove/Contradict that suppose $a_n$ and $b_n$ are monotonic sequences, therefore ${a_n} + {b_n}$ has a limit in the broad meaning of limits (a finite limit, or $\infty$, or $-\infty$).

I started solving by:

1.Since both sequences are monotonic, each of them has a broad limit (a finite one, or $+/-\infty$).

2.By using limits arithmetic, since ${a_n}, {b_n}$ has a broad limit, ${a_n} + {b_n}$ has a broad limit also (the same way I've shown before).

But this is incorrect.

So when is it possible showing that a sequence converges (also to $+/- \infty$) due to limits arithmetic rule?

Thank you.

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    $\begingroup$ I don't see where the contradiction happened. Can you be more specific about that? $\endgroup$ – user258700 Dec 13 '15 at 15:01
  • $\begingroup$ Counterexample: $a_n = 5n+(-1)^n$, $b_n = -5n$, both sequecnes are monotonic (and have a limit of +- infinity) but $a_n + b_n = (-1)^n$ is divergent. $\endgroup$ – Taru Dec 13 '15 at 15:08
  • $\begingroup$ I understand, but this isn't in contradiction with the initial statement. What I meant was: can you show where the contradiction to statement happened? $\endgroup$ – user258700 Dec 13 '15 at 15:19
  • $\begingroup$ @AhmedHussein In my previous question, I was told that if I have 3 sequences: $a_n, $b_n, a_n+b_n$ and two of them converges, then the third one converges also due to limits arithmetic rule. But this doesn't happen on this case, how come? $\endgroup$ – Taru Dec 13 '15 at 15:26
  • $\begingroup$ It's not true as you state it. I think you misinterpreted it. If $a_n$ and $b_n$ converge, then so does their sum; as an application, if $a_n$ converges and so does $a_n + b_n$, then $b_n$ converges because $b_n = (a_n + b_n) + (-a_n)$. Another application is that if $a_n$ converges and $b_n$ diverges, then their sum diverges because if it were convergent, then so would have been $(a_n + b_n) + (-a_n) = b_n$ which is not true. You can deduce many other things as well, but your examples do not fit. $\endgroup$ – user258700 Dec 13 '15 at 15:49

You must be careful about using the limit sum or product formula when dealing with infinities.

For example a limit of $\infty$ minus a limit of $\infty$ is ambiguous. So is a limit of 0 times a limit of $\infty$

In these situations you need to add or multiply the actual sequences to calculate the limit if it exists.

  • $\begingroup$ Well in class we've proved that you can use sum, and multiplication with infinite limits... (though with some restrictions...) $\endgroup$ – Taru Dec 13 '15 at 15:34
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    $\begingroup$ I think one of the restrictions is that at least one of the sequences must have a finite limit. $\endgroup$ – Ameet Sharma Dec 13 '15 at 15:38
  • $\begingroup$ Ok, one of the restrictions is that you cannot know about $\infty-\infty$. Maybe that's the clue. $\endgroup$ – Taru Dec 13 '15 at 15:40
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    $\begingroup$ That's right, that's an ambigrous situation. you're right that it does work in certain situations, but when infinities are involved there'll be exceptions that won't work. Another exception would be the product of a limit of 0 times a limit of $\infty$. This is also ambiguous, and you'll need to look the actual sequence product to calculate the limit. $\endgroup$ – Ameet Sharma Dec 13 '15 at 15:50
  • $\begingroup$ I've edited it. $\endgroup$ – Ameet Sharma Dec 13 '15 at 16:02

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