Sequence divergence test

Question

Could I argue that $\left(1+\frac{1}{n}\right)^{n^2}$ = $e ^n$ therefore the sequence diverges?

I am wondering if it is a legal move

Answer 1

The idea is correct, but that's not a proof. On the other hand, you can say that $(\forall n\in\mathbb N):\left(1+\frac1n\right)^n\geqslant2$ and that therefore$$(\forall n\in\mathbb N):\left(1+\frac1n\right)^{n^2}=\left(\left(1+\frac1n\right)^n\right)^n\geqslant2^n.$$ That's enough to prove that your sequence diverges.

Answer 2

The equation you have written is not correct. $(1+\frac 1 n)^{n^{2}} =((1+\frac 1 n)^{n})^{n} > c^{n}$ for $n$ sufficiently large if $1<c<e$ Since $c^{n} \to \infty$ the conclusion follows.

Answer 3

$\left(1+\frac{1}{n}\right)^{n^2} \lt e^n$ so that does not quite work

but $e^{n-1} \lt \left(1+\frac{1}{n}\right)^{n^2} \lt e^n$ for $n \gt 1$ and that would be enough if you could prove it

Answer 4

Option: Binomial expansion:

$(1+1/n)^{n^2} =$

$ 1+(n^2)(1/n)+....> n.$

Take the limit $n \rightarrow \infty$.

Answer 5

In this case, since no indetermination is generated in the process, you can actually just say that

$$ \lim \left(1+\frac 1n\right)^{n^2} = \left(\lim \left(1+\frac 1n \right)^n\right)^{\lim n}=e^{+\infty} =+\infty. $$