Show that $P(n) = \frac{3}{2} \times \frac{5}{4} \times ...\frac{2^n+1}{2^n}<e$

Question

The question says to use the arithmetic geometric inequality and the fact that $(1+\frac{1}{n})^n < e$. I have tried simply putting the terms into the AM-GM inequality which gets $P(n) < (1-2^{-n}+\frac{1}{n})^{n}$ but that is useless. Any help?

Answer 1

Hint: $\left(\dfrac 32\right)^2=\left(\dfrac {2+1}{2}\right)^2<e$

Do the same for other terms, what can you say about $e^{\frac 12}e^{\frac 14}\cdots e^{\frac 1{2^n}}$ ?

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In fact we can use the AM-GM inequality, here is how:

$\sqrt[n]{\dfrac 32\times\dfrac 54\cdots\times\dfrac{2^n+1}{2^n}}\le\dfrac 1n\left(\dfrac 32+\dfrac 54+\cdots\dfrac{2^n+1}{2^n}\right)=\dfrac 1n\left(n+\sum\limits_{k=1}^n\frac 1{2^k}\right)=\dfrac 1n\left(n+1-\dfrac 1{2^n}\right)$

Now just ignore the negative quantity and we get $\sqrt[n]{P(n)}\le\dfrac 1n(n+1)=1+\dfrac 1n$

Finally $P(n)\le\left(1+\dfrac 1n\right)^n<e$

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Remark: in the initial method we have $P(n)\le e^{\frac 12}e^{\frac 14}\cdots e^{\frac 1{2^n}}=e\,^{\sum\limits_{k=1}^n\frac 1{2^k}}=e^{1-\frac 1{2^n}}<e$

Due to the property of exponential function to transform products into sums, this is in fact equivalent to the AM-GM in disguise, so you should not be surprised (this is in fact how we prove the AM-GM in the first place, by convexity of exponential).