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While studying for my upcoming exams I came across a problem in the AM-GM section:

If $a_n = (1+\frac{1}{n})^{n}$ , $n \in \mathbb N$ then prove that:

$$2 < a_n < 4$$

Proving the lower bound is easy enough and I can prove the upper bound using the binomial expansion but I am not sure how to do this using the AM-GM inequality. Any help is much appreciated!

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This wasn't easy for me to prove. We take:

\begin{equation*} b_1 = b_2 = ... = b_{n-1} = 1,\ b_n = b_{n+1} = \frac12 \end{equation*} Then:

$$ \frac{\sum^{n+1}_{k=1}{b_k}}{n+1} > \sqrt[n+1]{\prod^{n+1}_{k=1}{b_k}} $$

$$ \frac{n}{n+1} > \sqrt[n+1]{\frac14} $$

$$ \left(\frac{n}{n+1}\right)^{n+1} > \frac14 $$

$$ \left(\frac{n+1}{n}\right)^{n+1} < 4$$

$$ \left(\frac{n+1}{n}\right)^n \left(\frac{n+1}{n}\right) < 4 $$

$$ \left(\frac{n+1}{n}\right)^n < 4 \left(\frac{n}{n+1}\right) $$

$$ \left(\frac{n+1}{n}\right)^n < 4 $$

$$ \left(1 + \frac1n \right)^n < 4 $$

$$ a_n < 4 $$

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  • $\begingroup$ Everything is fine just change the sequence you used for the AM-GM so some people don't get confused. $\endgroup$ – HeatTheIce Jan 1 '15 at 20:22
  • $\begingroup$ @SoulEater Fixed. $\endgroup$ – Veritas Jan 1 '15 at 20:33
  • $\begingroup$ This is very clever. $\endgroup$ – Ali Caglayan Jan 1 '15 at 21:59
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Here is a proof from N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563.

Let $a_n = (1+\frac1n)^n$ and $b_n = (1+\frac1n)^{n+1}$. We will prove that $a_n$ is an increasing sequence and $b_n$ is an decreasing sequence. Since $a_n < b_n$, this implies for any positive integers $n$ and $m$ with $m < n$ that $a_m < a_n < b_n < b_m$.

We use the AM-GM inequality in the form $$\left(\frac{v_1+v_2+...v_n}{n}\right)^n > v_1v_2...v_n$$ (all $v_i$ positive), with equality if and only if all the $v_i$ are equal (this allows us to avoid the use of n-th roots).

For $a_n$, consider $n$ values of $1+\frac{1}{n}$ and $1$ value of $1$. By the AM-GMI, $$\left(\frac{n+2}{n+1}\right)^{n+1} > \left(1+\frac1n\right)^n$$ $$\left(1+\frac1{n+1}\right)^{n+1} > \left(1+\frac1n\right)^n$$ $$a_{n+1}> a_n$$ $$a_n < a_{n+1}$$

For $b_n$, consider $n$ values of $1-\frac1n$ and $1$ value of $1$. By the AM-GMI, $$\left(\frac{n}{n+1}\right)^{n+1} > \left(1-\frac1n\right)^n$$ $$\left(\frac{n+1}{n}\right)^{n+1} < \left(\frac{n}{n-1}\right)^n$$ $$\left(1 + \frac1n\right)^{n+1} < \left(1 + \frac1{n-1}\right)^n$$ $$b_n < b_{n-1}$$ $$b_n > b_{n+1}$$

Therefore, $$a_n < b_n \le b_1 = \left(1 + \frac11\right)^2 = 4$$ $$a_n < 4$$

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