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Suppose $a_n \rightarrow +-\infty$ and $(b_n)$ is bounded. Show that $a_n+b_n \rightarrow +-\infty$. I tried this:

$|a_n|\rightarrow +-\infty$, so $|a_n+-\infty|<\epsilon$. It is also true that $|b_n|<M$, because $(b_n)$ was bounded. Now, if you add those together, you will get: $|a_n+-\infty|+|b_n|<\epsilon+M$ and therefor: $|a_n+b_n+-\infty|\le|a_n+-\infty|+|b_n|<\epsilon+M$. But, since M is fixed (say M was $2$), you can't get closer than $2$ to the limit (because $\epsilon$ must be negative in that case). Am I missing something?

Regards, Kevin

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I wouldn't advise you to add/subtract infinity until you'll have enough experience in this.

The strict proof is like this:

  1. suppose that $a_n\to+\infty$, so for any $E>0$ (here we are especially interested in large values of $E$) there exists $N(E)$ such that $a_n>E$ for all $n\geq N$.

  2. As you have written, there is a constant $M$ such that $|b_n|<M$ for all $n\geq0$, i.e. $$ -M<b_n<M. $$ We will use the left-hand side of the inequality.

  3. To prove that $a_n+b_n\to+\infty$ we should show that for any $E'$ there is $N(E')$ such that for all $n\geq N(E')$ holds $a_n+b_n>E'$.

  4. We can clearly do it: pick up any $E'$, then $a_n+b_n>a_n-M$ (see 2.), hence to make $a_n+b_n>E'$ we just need to make $a_n>E'+M$ for any $E'$ - and that will be sufficient (do you agree here?)

  5. Based on 1., we just take $N(E'+M)$ so $a_n>E'+M$ for all $n\geq N(E'+M)$, hence $$ a_n+b_n>E' $$ for all $n\geq N(E'+M)$ and hence $a_n+b_n\to +\infty$.

Could you please follow the same steps to prove the case when $a_n\to -\infty$?

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  • $\begingroup$ Ok, it was a little hard to follow for me, but now I get it. Just one question left: why do you pick N(E'+M), I thought you'd have to prove that $a_n>E \forall n \ge N \forall E$, because it should hold for every E? Else you can say $a_n=1>E$ or something like that. $\endgroup$ Sep 23 '11 at 11:44

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