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Given $$ y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1. $$ Show that $\lbrace y_n \rbrace$ is a decreasing sequence. Anyone can help ? I consider the ratio $\frac{y_{n+1}}{y_n}$ but I got stuck.

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    $\begingroup$ See this post. $\endgroup$ – David Mitra Feb 17 '13 at 13:26
  • $\begingroup$ @DavidMitra: thx for the link. $\endgroup$ – Idonknow Feb 17 '13 at 13:36
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Note that $$ \begin{align} \frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}} &=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\ &=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\ &=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\ &\ge\frac{n}{n+1}\left(1+\frac{n+2}{n(n+2)}\right)\\ &=1 \end{align} $$ Therefore, $$ \left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1} $$


Similarly, $$ \begin{align} \frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n} &=\left(\frac{n+2}{n+1}\right)^{n+1}\left(\frac{n}{n+1}\right)^n\\ &=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\ &=\frac{n+1}{n}\left(1-\frac1{(n+1)^2}\right)^{n+1}\\ &\ge\frac{n+1}{n}\left(1-\frac{n+1}{(n+1)^2}\right)\\ &=1 \end{align} $$ Therefore, $$ \left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n $$


Bernoulli's Inequality

In the preceding, we used Bernoulli's Inequality: for all $x\ge-1$ and non-negative integer $n$, $$ (1+x)^n\ge1+nx $$ This can be proven by induction:

Note that the inequality above is true for $n=0$.

Suppose that $x\ge-1$ and for a non-negative integer $n$, we have $$ (1+x)^n-nx\ge1 $$ Then $$ \begin{align} (1+x)^{n+1}-(n+1)x &=(1+x)^n-nx+x(1+x)^n-x\\ &\ge1+x((1+x)^n-1)\\ &\ge1 \end{align} $$ If $-1\le x\le0$, then both $x$ and $(1+x)^n-1$ are negative. If $x\ge0$, then both both $x$ and $(1+x)^n-1$ are positive. Therefore, if $x\ge-1$, $x((1+x)^n-1)\ge0$. This justifies the last inequality above.

Note that if $x\ne0$ and $n\ge1$, the last inequality is strict. Thus, for $x\ne0$ and $n\ge2$, we have $$ (1+x)^n\gt1+nx $$


Negative Exponents

Bernoulli's Inequality is also true for negative integer exponents. That is, for $x\gt-1$ and non-negative $n\in\mathbb{Z}$, $$ 1-nx\le(1+x)^{-n} $$ Suppose that $$ (1-nx)(1+x)^n\le1 $$ which is trivially true for $n=0$, and strictly true for $x\ne0$ and $n=1$. Then $$ \begin{align} (1-(n+1)x)(1+x)^{n+1} &=(1-nx)(1+x)^n-(n+1)x^2(1+x)^n\\ &\le1 \end{align} $$ Therefore, for all non-negative $n\in\mathbb{Z}$, $$ (1-nx)\le(1+x)^{-n} $$ where the inequality is strict for $x\ne0$ and $n\ge1$.

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  • $\begingroup$ In this answer, I extend Bernoulli's inequality to rational exponents. $\endgroup$ – robjohn Apr 16 '13 at 15:53
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I think it needs just some basic algebraic manipulations: $$\frac{y_{n}}{y_{n+1}}=\frac{(1+\frac{1}n)^{n+1}}{(1+\frac{1}{n+1})^{n+2}}$$ You can show that the latter fraction is equal to $$(1+\frac{1}{n^2+2n})^{n+1}\times\frac{1}{1+1/(n+1)}$$ But $$(1+\frac{1}{n^2+2n})^{n+1}\ge {1+1/(n+1)}$$ Note that if $x\ge-1$ then $(1+x)^n\ge 1+nx.$

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  • $\begingroup$ is there a typo in the first equation, the power of denominator? $\endgroup$ – Idonknow Feb 17 '13 at 13:37
  • $\begingroup$ Yes. I fixed it. $\endgroup$ – mrs Feb 17 '13 at 13:43
  • $\begingroup$ Good work! +1 ${}{}$ $\endgroup$ – Namaste Feb 18 '13 at 0:08
  • $\begingroup$ @amWhy: Thanks. I was the first here to reply. $\endgroup$ – mrs Feb 18 '13 at 3:57
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If you don't care about overkilling the Problem you can even do this with calculus, showing the derivative in n is negative. The derivative is $$ \frac{\left(\frac{1}{n}+1\right)^n (n+1) \left(n \log \left(\frac{1}{n}+1\right)-1\right)}{n^2}$$ and so we only need to show that $$1> n \log(1+\frac{1}{n})$$ By substitution $n=\frac{1}{x}$ we have $$\frac{\log(1+x)}{x}=\frac{\ln(1+x)-\ln(1)}{(x+1)-1}=\frac{1}{1+\xi}$$ with $\xi \in (0,x)$ (this is granted by the mean value theorem), and the last expression is less than 1.

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Use the first derivative test and prove the function

$$ f(x)=\left(1+\frac{1}{x}\right)^{x+1}\hspace{-6 mm},\qquad \quad x \geq 1, $$

is decreasing on $[1,\infty]$. That is prove $f'(x)<0$ on $[1,\infty]$.

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Lemma(1):

Let be $a,b\in\mathbb R^+$

$$\sqrt[n+1]{ab^n}\le \dfrac{a+bn}{n+1}$$

Proof:

$$\sqrt[n+1]{a\underbrace{bbb...b}_{n\;times}}\le \dfrac{a+b+b+b+...+b}{n+1}=\dfrac{a+bn}{n+1}\\ \Box.$$

Lemma(2):

Let be $x_n=\left(1+\dfrac1n \right)^{n}$ and $z_n=\left(1-\dfrac1n\right)^{n},\quad \forall n\neq 0\in\mathbb N,\quad x_n<x_{n+1}\quad and\quad z_n<z_{n+1}$

Proof: Use "Lemma(1)" and choose $a=1$ and $b=\left(1\pm\dfrac1n\right)$ $$\Longrightarrow$$ $$\sqrt[n+1]{\left(1\pm\dfrac1n\right)^{n}}<\dfrac{1+n\left(1\pm\frac1n\right)}{n+1}=1\pm\dfrac1n$$ $$\Longrightarrow$$ $$\left(1\pm\dfrac1n\right)^{n}<\left(1\pm\dfrac1n\right)^{n+1}$$$$\Box.$$

Theorem:

$y_n=\left(1+\dfrac1n\right)^{n+1},\quad \forall n\neq 0\in\mathbb N,\quad y_{n+1}<y_n$

Proof:

We know that $z_{n+1}<z_{n+2}\Longleftrightarrow \dfrac1{z_{n+1}}>\dfrac1{z_{n+2}}$ from "Lemma(2)"

$$y_n=\left(1+\dfrac1n\right)^{n+1}=\left(\dfrac{n+1}{n}\right)^{n+1}=\dfrac{1}{\left(\dfrac{n}{n+1}\right)^{n+1}}=\dfrac{1}{\left(1-\dfrac1{n+1}\right)^{n+1}}=\dfrac1{z_{n+1}}$$

$$y_n>y_{n+1}\Box.$$

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Hint: Try to use the A.M.- G.M. inequality for positive numbers.

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the derivative of $(1+1/n)^{(n+1)}$ is $(ln(1+1/n)-1/n)*(1+1/n)^{(n+1)}$ so if that is less than $0$ then $(1+1/n)^{(n+1)}$ is decreasing. $(1+1/n)^{(n+1)}$ is greater than $0$. if $ln(1+1/n)-1/n<0$ and $(1+1/n)^{(n+1)}>0$ then $(ln(1+1/n)-1/n)*(1+1/n)^{(n+1)}<0$. to show that $ln(1+1/n)-1/n<0$ you add 1/n to both sides and get $ln(1+1/n)<1/n$ let $x=1/n$ so then I am proving $ln(1+n)<n$ so $n+1<e^n$ by bernellies inequaility this is true for x is not 0, but the original function is undefined at x=0, so that doesn't matter. I am left with proving $(1+1/n)^{(n+1)}>0$ and this is all that I was able to do. can somebody finish that from there please?

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