# Prove that this sequence diverges to infinity.

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

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.

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"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? –  KitKat Nov 13 '13 at 22:09
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
"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. –  KitKat Nov 13 '13 at 22:11

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

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Can't believe I didn't see Bernoulli coming. Thanks! –  KitKat Nov 13 '13 at 22:17

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

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.

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How does this answer the question? –  Lord Soth Nov 13 '13 at 22:23
Show that $\forall k \in R$ $\exists N_k > 0$ such that $n > k$ $\forall n > N_k$ Take any $k \in R$ Then if $N_k = n$ you are done. –  Zhoe Nov 13 '13 at 22:31
Great, can you then show how it is done for this particular problem? –  Lord Soth Nov 13 '13 at 22:33
Not 100% certain of correctness, but that's how I'd approach it. –  Zhoe Nov 13 '13 at 23:21

An approach would be using the fact that:

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

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How would this approach work ? –  Amr Nov 13 '13 at 22:12
@user2896626 this lacks rigour. –  Amr Nov 13 '13 at 22:14
@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
@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
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 Nov 14 '13 at 0:19