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Please, help me to prove that $$x_n=\left(1+\frac{1}{n}\right)^{n+1}$$decreases. I know I must to prove that that $$\frac{x_n}{x_{n+1}}> 1$$

What to do next?

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marked as duplicate by robjohn Apr 19 '13 at 22:19

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This seems like a duplicate of this question. It is also shown there that $\left(1+\frac1n\right)^n$ is an increasing sequence. –  robjohn Apr 19 '13 at 22:19

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Write out $$ \frac{x_{n+1}}{x_n} = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( \frac{1+\frac{2}{n}}{\left(1+\frac{1}{n} \right)^2} \right)^n = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{\frac{1}{n^2}}{\left(1+\frac{1}{n} \right)^2} \right)^n = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{1}{\left(n+1 \right)^2} \right)^n = \left(1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3} \right) \left( 1 - \frac{1}{\left(n+1 \right)^2} \right)^n $$ we now have to use $(1-x)^n \leqslant \frac{1}{1+n x}$ for $0<x<1$ and $n \geqslant 1$: $$ \frac{x_{n+1}}{x_n} \leqslant \frac{1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3}}{ 1+\frac{n}{(n+1)^2} } < 1 $$

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Sasha, we have $x_n=\left(1+\frac{1}{n}\right)^{n+1}$ and $x_{n+1}=\left(1+\frac{1}{n+1}\right)^{n+2}$. From this, how can I get $\frac{x_{n+1}}{x_n} = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( \frac{1+\frac{2}{n}}{\left(1+\frac{1}{n} \right)^2} \right)^n$ ? –  Walter r Apr 18 '13 at 21:47
    
$$\frac{x_{n+1}}{x_n} = \frac{(n+2)^{n+2} \cdot n^{n+1}}{ (n+1)^{n+2} \cdot (n+1)^{n+1}} = \frac{\left(1+\frac{2}{n} \right)^{n+2} \cdot n^{n+2} \cdot n^{n+1}}{ \left(1+\frac{1}{n}\right)^{n+2} \cdot n^{n+2} \cdot \left(1+\frac{1}{n}\right)^{n+1} \cdot n^{n+1}} $$ Now simplify and regroup. –  Sasha Apr 18 '13 at 22:08

Let $f(x) = (1+\frac{1}{x})^{x+1}$. In the following, we show that $f(x)$ is decreasing for $x>0$. We have $f(x) = \exp[(x+1)\log(1+\frac{1}{x})]$ and it is thus sufficient to show that $g(x) = (x+1)\log(1+\frac{1}{x})$ is decreasing. Indeed, $$g'(x) = \log\left(1+\frac{1}{x}\right) - \frac{1}{x} < \frac{1}{x} - \frac{1}{x} = 0,$$ as $\log(1+y)<y,\,\forall y> 0$.

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Usually one studies this sequence in order to introduce the exponential function and its inverse, the logarithm. Decreasing of this sequence can be proven by more elementary methods. The book "Analysis by its history" starts this way. –  Sasha Apr 18 '13 at 18:43

This can be shown using AM $\ge$ GM or using Bernoulli's ineqaulity.

1) Using AM $\ge$ GM

Take one copy of $1$, and $n$ copies of $1 - \frac{1}{n}$, ($n \gt 1$) to get

$$ \frac{1 + n(1 - \frac{1}{n})}{n+1} \gt \left(1-\frac{1}{n}\right)^{n/(n+1)}$$

which simplies to

$$ \left(\frac{n}{n+1}\right)^{n+1} \gt \left(\frac{n-1}{n}\right)^n$$

Taking the reciprocals, yields

$$ \left(1 + \frac{1}{n}\right)^{n+1} \lt \left(1 + \frac{1}{n-1}\right)^n $$

2) Using (generalized) Bernoulli's inequality

$$(1 + x)^r \lt 1 + rx $$ for the case when $x \gt -1$, $x \neq 0$, and $0 \lt r \lt 1 $

Here we set $x = -\frac{1}{n}$, $r = \frac{n}{n+1}$ and obtain the result easily.


For a similar problem and multiple proofs of that (including AM,GM and bernoulli's inequality), see: Proving : $ \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} $

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As far as I know, this was first done in 1951: N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 –  marty cohen Apr 19 '13 at 1:07
    
@martycohen: Thank you for the reference! –  Aryabhata Apr 19 '13 at 1:10

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