# Using sequence limits arithmetic rules to prove convergence of a sequence

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

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

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.

• I don't see where the contradiction happened. Can you be more specific about that? – user258700 Dec 13 '15 at 15:01
• 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. – Taru Dec 13 '15 at 15:08
• 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? – user258700 Dec 13 '15 at 15:19
• @AhmedHussein In my previous question, I was told that if I have 3 sequences: $a_n,$b_n, a_n+b_n$and two of them converges, then the third one converges also due to limits arithmetic rule. But this doesn't happen on this case, how come? – Taru Dec 13 '15 at 15:26 • It's not true as you state it. I think you misinterpreted it. If$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$which is not true. You can deduce many other things as well, but your examples do not fit. – user258700 Dec 13 '15 at 15:49 ## 1 Answer You must be careful about using the limit sum or product formula when dealing with infinities. For example a limit of$\infty$minus a limit of$\infty$is ambiguous. So is a limit of 0 times a limit of$\infty$In these situations you need to add or multiply the actual sequences to calculate the limit if it exists. • Well in class we've proved that you can use sum, and multiplication with infinite limits... (though with some restrictions...) – Taru Dec 13 '15 at 15:34 • I think one of the restrictions is that at least one of the sequences must have a finite limit. – Ameet Sharma Dec 13 '15 at 15:38 • Ok, one of the restrictions is that you cannot know about$\infty-\infty$. Maybe that's the clue. – Taru Dec 13 '15 at 15:40 • That's right, that's an ambigrous situation. you're right that it does work in certain situations, but when infinities are involved there'll be exceptions that won't work. Another exception would be the product of a limit of 0 times a limit of$\infty\$. This is also ambiguous, and you'll need to look the actual sequence product to calculate the limit. – Ameet Sharma Dec 13 '15 at 15:50
• I've edited it. – Ameet Sharma Dec 13 '15 at 16:02