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Prove that the sequence $$b_n=\left(1+\frac{1}{n}\right)^{n+1}$$ Is decreasing.

I have calculated $b_n/b_{n-1}$ but it is obtain: $$\left(1-\frac{1}{n^2}\right)^n \left(1+\frac{1}{n}\right)^n$$ But I can't go on.

Any suggestions please?

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marked as duplicate by Martín-Blas Pérez Pinilla, Davide Giraudo, Mark Fantini, nbubis, user147263 Oct 23 '14 at 11:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$y=\left(1+\frac{1}{x}\right)^{x+1}$$

$$\ln y=({x+1})\cdot\ln\left(1+\frac{1}{x}\right)$$

$$y'\frac{1}{y}=\ln\left(1+\frac{1}{x}\right)+(x+1)\cdot \frac{1}{1+\frac{1}{x}}\cdot\left(-\frac{1}{x^2}\right)$$

$$y'=\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{x}\right)\cdot\left(1+\frac{1}{x}\right)^{x+1}$$

$$\Rightarrow y'<0$$

Hence $y$ is decreasing

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Consider $f(x) = (x+1)\ln (x+1) - (x+1)\ln x$ on $x \geq 1$, we have:

$f'(x) = \ln\left(1 + \dfrac{1}{x}\right) - \dfrac{1}{x} < 0$ because it is true that $e^r \geq 1 + r$, and apply this for $r = \dfrac{1}{x} > 0$. From this the answer follows.

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Why don't you change your goal to prove that $b_n-b_{n+1}>0$?

$b_n-b_{n+1}=(1+1/n)^n-(1+1/n)^n(1+1/n)=(1+1/n)^n(1-1-1/n)=(1+1/n)^n(-1/n)$

Since $n$ is a natural number, $-1/n < 0$. Since $1>0, 1/n>0$, we have $1+1/n>0$. It would be nice if you can use $(1+1/n)^n >0$ directly based on $1+1/n>0$. If not, just do a simple proof by induction.

Then you will have your $b_n-b_{n+1}>0$, i.e., $b_n> b_{n+1}$, meaning that $b_n$ is decreasing.

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  • 1
    $\begingroup$ Please use MathJax. It makes answers much more readable. Also, notice that this question is a duplicate. $\endgroup$ – robjohn Oct 23 '14 at 6:57
  • $\begingroup$ Your statement that "bn-bn+1>0" seems to say "1>0" which is true, but trivial. I am sure this is not what you meant; I am sure what you meant was $b_n-b_{n+1}\gt0$. $\endgroup$ – robjohn Oct 23 '14 at 7:00
  • $\begingroup$ Now that OC-Sansoo has kindly converted your post to MathJax, it appears that you are mistakenly saying that $b_n=(1+1/n)^n$ and $b_{n+1}=(1+1/n)^{n+1}$. One or both of these is wrong. $\endgroup$ – robjohn Oct 23 '14 at 7:05
  • $\begingroup$ Your worry is valid. Let me redo the calculation $\endgroup$ – QmmmmLiu Oct 23 '14 at 7:13
  • $\begingroup$ Your answer supposedly shows that $b_n-b_{n+1}\lt0$ which means $b_n$ is increasing. This is not right. $\endgroup$ – robjohn Oct 23 '14 at 7:13
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Since the inequality holds for any positive numbers:$$\sqrt[n]{{a_1}{a_2}\cdots {a_n}}\ge \frac{n}{a_1^{-1}a_2^{-1}\cdots a_n^{-1}},$$we have:$$\sqrt[n+2]{\underbrace{\frac{n+1}n\frac{n+1}n\cdots \frac{n+1}n}_{n+1\text{ times}}\cdot 1}\ge \frac{n+2}{\frac n{n+1}+\frac n{n+1}+\cdots +\frac n{n+1}+1}.$$Hence,$$\left(1+\frac 1n\right)^{n+1} \ge \left(1+\frac 1{n+1}\right)^{n+2}.$$

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Hint: using the tools of differential calculus, study the function $$x\longmapsto \left(1+\frac{1}{x}\right)^{x+1}$$ for $x>0$.

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My 2¢: consider the function defined apriori for $x>0$ $$f(x)=\log(1+x)\cdot (\frac{1}{x}+1)= \frac{\log(1+x)\cdot (1+x)}{x}$$

$f$ extends analytically to $(-1, \infty)$, and continuously to $[-1, \infty)$. We have $f(-1)=0$ and $f(0)=1$.

We calculate: $$f'(x)= \frac{x - \log(1+x)}{x^2}$$ so $f'(x)>0$ for $x> -1$, $x \ne 0$ and so $f$ strictly increasing on $(-1, \infty)$. Now consider the decreasing sequence of values $\frac{1}{n}$ for $x$.

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  • $\begingroup$ It might be useful to point out that this corresponds to the question by $x=1/n$ and that since $f(x)$ is increasing in $x$, $f(1/n)$ is decreasing in $n$. $\endgroup$ – robjohn Oct 23 '14 at 8:18
  • $\begingroup$ @robjohn: Correct. $\endgroup$ – Orest Bucicovschi Oct 23 '14 at 8:44

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