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As $n$ goes to infinity: $\lim_{n\to\infty} a_n = 0$ , $\lim_{n\to\infty}b_n = \infty$

I need to prove that $\lim_{n\to\infty}a_n^{b_n} = 0$. Is it enough to say that from a curtain point $|a_n| < 1$ and thus $lim_{a_n^{b_n}} = 0$ because $b_n$ is greater than $1$ from a curtain point?

DarkLeader
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  • SO, $a_n$ could be negative and $b_n$ could be irrational? Yet you still talk about $a_n^{b_n}$ ? – GEdgar Dec 23 '19 at 21:23
  • I like your idea! So what you’re saying is that eventually, $|a_n| < \epsilon$ for large enough $n$, and $b_n>1$ for large enough $n$, so $(|a_n|)^{b_n} < \epsilon^{b_n} < \epsilon$ for large enough $n$. – layman Dec 23 '19 at 21:28
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    @layman yeah it just makes sense to me. is it a valid proof though? – DarkLeader Dec 23 '19 at 21:29
  • @darkleader it sounds valid to me! And GEdgar’s answer below agrees, though it’s slightly different, but same idea. You just have to clean it up a bit by using the “for every $\epsilon$, there exists an $N$” language but the main idea is there! – layman Dec 23 '19 at 21:31
  • @layman I can write the whole thing out with epsilon and find n>N and all that but I'm looking a shorcut I guess – DarkLeader Dec 23 '19 at 21:32
  • @darkleader I think it’s hard to find a shortcut with these types of proofs. Typically they all have the $\epsilon-N$ stuff. If you’re having trouble incorporating that stuff into the proof see GEdgar or infinitylord below. If you’re still confused let me know, I can help, too. – layman Dec 23 '19 at 21:34
  • @layman the instructor said something about shortcuts of this type but I can't remember if he said that they will accept it or not. thanks anyways – DarkLeader Dec 23 '19 at 21:36

3 Answers3

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Fix $\epsilon > 0$. Since $a_n \to 0$, there exists an $N_a > 0$ such that $|a_n| < \min \{1, \epsilon \}$ whenever $n > N_a$. Moreover, $b_n \to \infty$ implies that there exists an $N_b > 0$ such that $b_n > 1$ whenever $n > N_b$

Now consider $N = \max \{ N_a, N_b \}$.

infinitylord
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  • thanks bud, but is my solution correct? I could've written this all out just wanted to know if my shortcut is valid. thanks again – DarkLeader Dec 23 '19 at 21:34
  • @DarkLeader There are some typos (forgetting the limit symbol for ${a_n}^{b_n}$, writing $a_n < |1|$) but yes your answer is in spirit the same as mine. I just formalized it. – infinitylord Dec 23 '19 at 21:42
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    thank you! much appreaciated – DarkLeader Dec 23 '19 at 21:43
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Let $\epsilon>0$ be given. There is $N$ so that for all $n \ge N$ we have $|a_n|<\epsilon$ and $b_n \ge 1$. Note: $a_n < 0$ is allowed, so $a_n^{b_n}$ could be complex (not real), and could have multiple values. But still, for $n \ge N$, we have $$ \big|a_n^{b_n}\big| = |a_n|^{b_n} \le |a_n|^1 = |a_n| <\epsilon $$

infinitylord
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GEdgar
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  • thanks bud, but is my solution correct? I could've written this all out just wanted to know if my shortcut is valid. thanks again – DarkLeader Dec 23 '19 at 21:37
  • Our answers were similar, and I accidentally edited yours thinking it was mine. I've undone those changes now. Sorry about that. – infinitylord Dec 23 '19 at 21:38
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I assume $a_n,b_n>0, (a_n)^{b_n}=e^{b_nln(a_n)}$

Since $lim_na_n=0, lim_nln(a_n)=-\infty, lim_nb_nln(a_n)=-\infty$ since there exists $N:b_n>1, n>N$

Tsemo Aristide
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