# Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence

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.

• See this post. – David Mitra Feb 17 '13 at 13:26
• @DavidMitra: thx for the link. – Idonknow Feb 17 '13 at 13:36

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 $n\ge1$.

• In this answer, I extend Bernoulli's inequality to rational exponents. – robjohn Apr 16 '13 at 15:53

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 $(1+x)^n\ge 1+nx.$

• is there a typo in the first equation, the power of denominator? – Idonknow Feb 17 '13 at 13:37
• Yes. I fixed it. – mrs Feb 17 '13 at 13:43
• Good work! +1 ${}{}$ – Namaste Feb 18 '13 at 0:08
• @amWhy: Thanks. I was the first here to reply. – mrs Feb 18 '13 at 3:57

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.

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 $[1,\infty]$.

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)"

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

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

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

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 $$ln(1+n) so $$n+1 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?