Proving this sequence is unbounded [duplicate]

I need to prove that the following sequence is unbounded:

$$Z_n := \left(1+\frac{1}{n}\right)^{n^2}$$ for $$n\in \mathbb N$$.

I know that I essentially need to show that $$\forall M \in\mathbb{R}$$ there exists $$z_N \in Z$$ such that $$M < z_N$$, i.e., $$M < \left(1+\frac{1}{N}\right)^{N^2}, N\in\mathbb{N}$$ But frankly, I'm lost as to how to prove that.

I was able to prove that $$(1+\frac{1}{n})^{n}$$ was bounded above by say, 3, by using the fact that $$\left(1+\frac{1}{n}\right)^{n} = \sum_{k=0}^n\binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^n\frac{1}{k!}(1)\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\dots\left(1-\frac{k-1}{n}\right)$$ and then showing that $$\sum_{k=0}^n\frac{1}{k!}(1)\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\dots\left(1-\frac{k-1}{n}\right) < \sum_{k=0}^n\frac{1}{2^k} < 3 \text{ (telescoping sum)}$$ But the same method does not work for the sequence $$Z_n$$. Any ideas? (Please have mercy)

Thanks!

• $(1+\frac{1}{n})^n$ approaches e as n approaches infinity, so your sequence should approach $e^n$, which is unbounded. I'm not sure how to make this rigorous, but hopefully it is helpful – Seth Nov 8 '18 at 3:14
• More specifically, $(1 + \frac{1}{n})^n$ INCREASES to $e,$ in particular it is larger than $2$ for $n > 1$ – Will Jagy Nov 8 '18 at 3:18
• $(1+\frac{1}{n})^{n^2}$ = ($1 + n^2*\frac{1}{n}$ + ... +$\frac{1}{n^{n^2}}$) > (1 + n) – Steve B Nov 8 '18 at 4:04

We prove that $$(1+1/n)^n \geqslant 2 [n \in \mathbb N^*]$$, and $$Z_n$$ is therefore unbounded.
By the binomial theorem, $$\left(1 + \frac 1n\right)^n = \sum_0^n \binom n k \frac 1{n^k} \geqslant \sum_0^1 \binom n k \frac 1{n^k} = \binom n 0 + \binom n1 \cdot \frac 1 n = 1 + 1 =2,$$ hence $$Z_n \geqslant 2^n,$$ and $$(2^n)_1^\infty$$ is unbounded.
Bernoulli inequality says that $$\left(1+\frac{1}{n}\right)^{n^2}\geq 1+\frac{n^2}{n}=n+1$$ and we are done.