2

$\lim_{n\to \infty} (1 +\frac{1}{n}\tag{displayed})^{n^2} = \infty$

I don't know how to tackle this one. Knowing that it diverges to infinity and thus does not have an upper bound, should I try to prove that it's an unbounded subsequence, if so how? Is that sufficient to show that $\infty$ is the limit?

Any help would be appreciated.

Nhat
  • 1,354
  • 2
    "Converges to infinity" is quite the oxymoron. – Ron Gordon Nov 13 '13 at 22:06
  • English is not my first language and I actually couldn't think of a n appropriate way to put it. What's the right way? – Nhat Nov 13 '13 at 22:09
  • 2
    You could say diverges to infinity. For that matter, you could say converges to infinity in the extended reals. Or you could be very informal and just say that it blows up. – Brian M. Scott Nov 13 '13 at 22:10
  • 2
    "Diverges." Sorry for the sarcasm. – Ron Gordon Nov 13 '13 at 22:10
  • Thanks. No problem, didn't even catch it before you mentioned it. – Nhat Nov 13 '13 at 22:11
  • Dear @RonGordon, what is wrong with "Converges to infinity"? It is more precise than "diverges"; not all divergent sequences converge to infinity! – Bruno Joyal Nov 13 '13 at 22:37
  • @BrunoJoyal, I'm trying to understand that. So, divergent simply means that limit does not extend, but converging to infinity means that it does converge in extended real systems. Am I right? – sprajagopal Aug 20 '14 at 05:37

3 Answers3

4

Hint: $$(1+\frac{1}{n})^{n^2}\geq 1+n^2(\frac{1}{n})$$

Amr
  • 20,030
2

$$\lim_{n\to \infty} a_n = + \infty$$

A sequence diverges to $+ \infty$, if $\forall k \in \mathbb{R}$, $\exists N_k \gt 0$ such that $a_n \gt k$, $\forall n > N_k$.

Take any $k \in R$

Using (from above) $$(1+\frac{1}{n})^{n^2}\geq 1+n^2(\frac{1}{n})$$

$$= 1 + n^2/n$$

$$=1 + n > n > k$$

$$\iff n > k$$

Take $N_k = k$ are you are done.

Zhoe
  • 2,415
-1

An approach would be using the fact that:

$$ \lim_{x \to \infty} \left(1+\frac{1}{n}\right)^n = e $$

Stefan Hamcke
  • 27,733
Paul92
  • 470
  • How would this approach work ? – Amr Nov 13 '13 at 22:12
  • Yes, I know about that. Sorry, if it wasn't obvious. – Nhat Nov 13 '13 at 22:13
  • His limit would be equal with lim e^n, as n->infinity – Paul92 Nov 13 '13 at 22:13
  • 1
    @user2896626 this lacks rigour. – Amr Nov 13 '13 at 22:14
  • Why? Can you explain, please? – Paul92 Nov 13 '13 at 22:15
  • @user2896626 You seem to be assuming the following statement: if $b_n$ converges to B, then $\lim_{n\rightarrow \infty} b_n^n=\lim_{n\rightarrow \infty} B^n$. While it may seem intuitive, this does not suffice to infer that it is true. We need a proof that it is true (In case it's realy true) – Amr Nov 13 '13 at 22:18
  • 1
    @user2896626 As a counterexample to the fact that you seem to assume: Set $b_n=1+\frac{1}{n}$. Then $b_n$ converges to $1$. However: $e=\lim_{n\rightarrow \infty} (1+\frac{1}{n})^n=\lim_{n\rightarrow \infty} b_n^n\not=\lim_{n\rightarrow} B^n=\lim_{n\rightarrow} 1^n =1$ – Amr Nov 13 '13 at 22:22
  • I suggest you have a look at "real analysis". I liked it a lot when I first learnt it, I also expect that you will like it. – Amr Nov 13 '13 at 22:24
  • I wish someone directed me to read analysis when I was wasting my time in calculus – Amr Nov 13 '13 at 22:30
  • No. I don't agree with the step where you take n out. Let me remind you that I gave you a counterexample in one of my previous comments. – Amr Nov 13 '13 at 22:33
  • @Amr, your counterexample is wrong, because lim 1^n = 1^infinity which is not equal to 1 (it's an indetermination). – Paul92 Nov 13 '13 at 22:34
  • @Amr, also, I hope you agree that lim(a * b) = lim a * lim b. Using induction, for lim(a^n) we get (lim a)^n. With n->infinity, we can take it's limit to compute it, so it becomes lim( (lim a)^n) – Paul92 Nov 13 '13 at 22:36
  • @user2896626 $\lim_{n\rightarrow} 1^n$ is just the limit of the sequence that consists of ${1,1^2,1^3,...}$ which is just a constant sequence. A constant sequence has converges to the constant value. you might not have a proof for it, but you can feel it – Amr Nov 13 '13 at 22:36
  • @Amr, I don't agree. Look in this table: http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms – Paul92 Nov 13 '13 at 22:38
  • Firstly, Do you still want to assert that for every sequence $b_n$ that converges to $B$, we have $\lim_{n\rightarrow \infty} b_n^n=\lim_{n\rightarrow \infty} B^n$ ? – Amr Nov 13 '13 at 22:48
  • Yes, I do, as far as the lim B^n exists. – Paul92 Nov 13 '13 at 22:52
  • But you don't agree that $\lim_{n\rightarrow \infty} 1^n$ exists ? – Amr Nov 13 '13 at 22:54
  • No. As the wikipedia post mentions, it is a indetermination, like 0^0. – Paul92 Nov 13 '13 at 22:56
  • OK. Ask on math exchange here the following question: What is $\lim_{n\rightarrow \infty} 1^n$ ? and I expect with a very high degree that you will get answers telling you that its one and its value is one. – Amr Nov 13 '13 at 22:59
  • But that's not a good argument, but just a thing that I said to make you trust me. I did not argue about intermediates because I dont know their definition (I also doubt that you know their definition). However, I know the defintion of a limit , I know what it exactly is. Thus, I can argue with you about limits. In particular I can argue with you about the value of $\lim_{n\rightarrow \infty} 1^n$. – Amr Nov 13 '13 at 23:02
  • 1
    @user2896626 The fact that $1^n$ results in an indeterminate form does not mean that its limit does not exist. In fact, $\lim_{n\rightarrow\infty} 1^n = 1$. – Lord Soth Nov 13 '13 at 23:04
  • Finally, I advice you to study analysis. I think you will like it. Your arguments are similar to the ones I made when I was studying calculus. In case you don't trust me (which is totally OK, as you don't know me) you may have the opinion of other people here on the site. Just ask a question and look at the answers and their votes – Amr Nov 13 '13 at 23:04
  • 1
    While I agree with @Amr 's obejction that, in general, if $b_n\rightarrow B$, this does not guarantee that $\lim_{n\rightarrow \infty} b_n^n = \lim_{n\rightarrow \infty} B^n$, in this particular case, it can be made to work. Namely, since $(1+\frac{1}{n})^n\rightarrow e$, for large enough $n$, $(1+\frac{1}{n})^n > 2$. Then, for large enough $n$, $(1+\frac{1}{n})^{n^2} = [(1+\frac{1}{n})^n]^n > 2^n$, so diverges. – Jason DeVito - on hiatus Nov 14 '13 at 00:19
  • @JasonDeVito Sure. – Amr Nov 14 '13 at 09:56