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This is a part of a larger proof I am doing by induction for an exercise, but I've gotten stuck on this part: $$\forall n \in \mathbb{N}, n > 1 \implies n \geq (1+\frac{1}{2n})^n$$

Any pointers would be appreciated.

Also I have no idea what to tag this as.

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    $\begingroup$ Looks like you only need to show it for $n=2$, because for $n\geq3$ we have $(1+\frac{1}{2n})^n<(1+\frac{1}{n})^n<e<3$. $\endgroup$ Sep 25, 2014 at 8:33

4 Answers 4

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${{\left( 1+\frac{1}{2(n+1)} \right)}^{n+1}}<{{\left( 1+\frac{1}{2n} \right)}^{n+1}}={{\left( 1+\frac{1}{2n} \right)}^{n}}\left( 1+\frac{1}{2n} \right)<n\left( 1+\frac{1}{2n} \right)=n+\frac{1}{2}<n+1$
using induction.

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It's a well known inequality that $$(1+x)^r \le e^{rx}, r > 0$$

Applying it here, $$\left(1+\dfrac{1}{2n}\right)^n \le e^{\frac{n}{2n}} = e^{\frac12} \simeq 1.648 < n\ \ \ \ \forall n > 1,\ \ n \in \mathbb N $$

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$n=2 \implies n > 1.5625 = (1+\dfrac{1}{2n})^n$

$n>2 \implies n > e > (1+\dfrac{1}{n})^n > (1+\dfrac{1}{2n})^n$

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Note that $$\left(1+\frac1{2n}\right)^n=\sum_{k=0}^n{n\choose k}\frac1{(2n)^k} $$ The first summand ($k=0$) is $1$ and ${n\choose k+1}\frac1{(2n)^{k+1}}\le \frac n{2n}\cdot {n\choose k}\frac1{(2n)^k} $ so that we get the estimate that the right hand side is $\le 1+\frac12+\frac14+\ldots=2$.

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