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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!

Community
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Eli
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  • $(1+\frac{1}{n})^n$ 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 08 '18 at 03:14
  • More specifically, $(1 + \frac{1}{n})^n$ INCREASES to $e,$ in particular it is larger than $2$ for $n > 1$ – Will Jagy Nov 08 '18 at 03:18
  • $(1+\frac{1}{n})^{n^2}$ = ($1 + n^2*\frac{1}{n}$ + ... +$\frac{1}{n^{n^2}}$) > (1 + n) – Steve B Nov 08 '18 at 04:04

2 Answers2

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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.

xbh
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Bernoulli inequality says that $$\left(1+\frac{1}{n}\right)^{n^2}\geq 1+\frac{n^2}{n}=n+1$$ and we are done.

Paramanand Singh
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