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I want to show that the infimum of the set of all numbers of the form $\big(1+\frac{1}{n}\big)^n$ where $n=1,2,3,...$ is $2$.

I thought I can prove this if I can show that $$\frac{a_{n+1}}{a_n} > 1 \quad \forall n$$ where $a_n=\big(1+\frac{1}{n}\big)^n$

$$\frac{a_{n+1}}{a_n}=\frac{ \bigg(1+ \frac{1}{n+1}\bigg)^{n+1} }{ \bigg(1+ \frac{1}{n}\bigg)^{n} } = \frac{ \bigg( \frac{n+2}{n+1}\bigg)^{n+1} }{ \bigg(\frac{n+1}{n}\bigg)^{n} } = \frac{ \bigg( \frac{n+2}{n+1}\bigg)^{n} }{ \bigg(\frac{n+1}{n}\bigg)^{n} } \bigg( \frac{n+2}{n+1}\bigg) = \bigg( \frac{(n+2)n}{(n+1)(n+1)}\bigg)^{n}\bigg(\frac{n+2}{n+1}\bigg) = \bigg(\frac{n^2+2n}{n^2+2n+1}\bigg)^{n}\bigg(\frac{n+2}{n+1}\bigg)$$

The first ratio is less than $1$, while the second one is greater than $1$.

I'm stuck at this point. Someone, please help.

Thuliyan
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    The link above is not exactly a duplicate, but it will certainly answer your question. – Mittens Jul 06 '21 at 18:44
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    To finish it the way you're trying: Note the inequality $$(1 - x)^n \geqslant 1 - nx$$ for $n \in \Bbb N$ and $x \in [0, 1]$. So the first term in your final product is $$\geqslant 1 - \frac{n}{n^2 + 2n + 1} = 1 - \frac{n}{(n + 1)^2}.$$ – Aryaman Maithani Jul 06 '21 at 18:47

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