how to prove that $(1 + \frac{1}{n})^{n+1}$ is decreasing?

Question

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 $$\frac{x_n}{x_{n+1}}> 1$$

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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

score 3 · Accepted Answer · answered Apr 18 '13 at 18:38

3

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 $n \geqslant 1$: $$ \frac{x_{n+1}}{x_n} \leqslant \frac{1+\frac{n}{(n+1)^2} - \frac{1}{(n+1)^3}}{ 1+\frac{n}{(n+1)^2} } < 1 $$

answered Apr 18 '13 at 18:38

Sasha

70,631

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

score 1 · Answer 2 · answered Apr 18 '13 at 18:40

1

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 $\log(1+y)<y,\,\forall y> 0$.

answered Apr 18 '13 at 18:40

Lord Soth

7,750
20
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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

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

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} $

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 01:07

how to prove that $(1 + \frac{1}{n})^{n+1}$ is decreasing?

3 Answers3