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I cannot prove that the function $$f (n) = \left(1 + \frac1n\right) ^ {n + 1},$$ defined for every positive integer $n$, is strictly decreasing in $n$. I already tried to prove by induction and also tried to prove by calculating the difference between $f (n + 1)$ and $f (n)$. I need help.

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May be this help:

$f_n/f_{n+1}=\frac{(1+1/n)^{n+1}}{(1+1/(n+1))^{n+2}}=(\frac{1+1/n}{1+1/(n+1)})^{n+1}\times\frac{1}{1+1/(n+1)}=(1+1/(n^2+2n))^{n+1}\times\frac{1}{1+1/(n+1)}$

But

$$(1+1/(n^2+2n))^{n+1}>1+\frac{n+1}{n^2+2n}>1+\frac{n+1}{(n+1)^2}=1+1/(n+1)$$

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$$f(n)=\exp{\left((n+1)\ln{\left(1+\frac1n\right)}\right)}$$ with $f(n)>0$. Hence $$f'(n)=f(n)\left(\ln{\left(1+\frac1n\right)-\frac1n}\right)$$ Now, use the inequality $\ln(1+x)<x$ for $x>-1$ with $x=\frac1n$ to conclude that $f'(n)<0$.

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    $\begingroup$ Are you serious? Why would this problem be presented after calculus, the exponential function and the logarithm? $\endgroup$
    – Git Gud
    Dec 25 '16 at 13:17
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    $\begingroup$ I agree with @GitGud It's hard to be sure, but that problem seems a lot more likely as a preparation for introducing $e$ than anything else, in any case something at the beginning of studying limits. As such that answer is not useful. // Having looked at OP's profile, maybe not. But it still seems way overkill to do it like this. $\endgroup$
    – quid
    Dec 25 '16 at 13:20
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    $\begingroup$ @quid Ok, fair enough. Now, that I see it again, I seem to agree with you, the OP must be looking for a much different answer. I will leave my answer anyway, just in case. $\endgroup$
    – Jimmy R.
    Dec 25 '16 at 13:24
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    $\begingroup$ @quid (1) The expression with $n$ in the exponent is used to prepare for $e$. Not $n+1$. Small but important difference. (2) It's not overkill. Computing the derivative to show whether or not a function is increasing or decreasing is a standard and simple method to solve these types of problems. $\endgroup$ Dec 25 '16 at 13:39
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    $\begingroup$ @MathematicsStudent1122 re 1, if you care to check say the Wikipedia page on e you'll see this one mention there too. The point to have a decreasing upper bound for e, in addition to an increasing lower bound for e, furnished by the sequence you mentioned. One can nicely squeeze e between these two sequences. re 2, overkill or not, it uses quite a bit of theory that may not be available to somebody having to solve this problem. But yeah likely it does not hurt to have this answer in addition to a different one. $\endgroup$
    – quid
    Dec 25 '16 at 13:56
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Take the fraction $$ \frac{f(n)}{f(n-1)}=\left(\frac{n+1}{n}\right)^{n+1}\bigg/\left(\frac{n}{n-1}\right)^n=\left(\frac{(n-1)(n+1)}{n^2}\right)^n\frac{n+1}{n}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}. $$ Using the induction, show that for $n\ge2$ and $x>-1$, $x\ne0$ (if $x=0$ then there would be a trivial equality) $(1+x)^n>1+nx$. For $n=2$ we have $(1+x)^2=1+2x+x^2>1+2x$. For $n+1$: $(1+x)^{n+1}=(1+x)^n(1+x)>(1+nx)(1+x)=1+(n+1)x+nx^2>1+(n+1)x$.

Using this inequality one can conclude that $$ \left(1+\frac{1}{n^2-1}\right)^n>1+n\frac1{n^2-1}=\frac{n^2+n-1}{n^2-1}, $$ and $$ \frac{f(n)}{f(n-1)}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}<\frac{n^2-1}{n^2+n-1}\frac{n+1}{n}=\frac{n^3+n^2-n-1}{n^3+n^2-n}<1. $$ Therefore, $f(n)$ is decreasing in $n$.

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Use Binomial Theorem,

$\left(1+\frac1n\right)^{n+1}>\left(1+\frac1{n+1}\right)^{n+2}=\left(1+\frac1{n+1}\right)^{n+1}\left(1+\frac1{n+1}\right)$

$\iff \left(1+\frac1n\right)^{n+1}\Big/\left(1+\frac1{n+1}\right)^{n+1}>\left(1+\frac1{n+1}\right)$

but LHS is,

$\left(1+\frac1{n(n+2)}\right)^{n+1}=1+\sum\limits_{k=1}^{n+1}\binom{n+1}{k}\left(\frac1{n(n+2)}\right)^k>1+\frac{n+1}{n(n+2)}>1+\frac{1}{n+1}$

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For $n>1$, we apply the AM-GM inequality to the numbers $(1-1/n)$ taken $n$ times and $1$. Then $$\underbrace{\frac{n(1-\frac{1}{n})+1}{n+1}}_{\text{Arithmetic mean}}> \underbrace{\left(\left(1-\frac{1}{n}\right)^n\cdot 1\right)^{1/(n+1)}}_{\text{Geometric mean}}$$ that is $$\frac{n^{n+1}}{(n+1)^{n+1}}> \frac{(n-1)^{n}}{n^{n}},$$ which is equivalent, to $$f(n-1)=\left(1+\frac{1}{n-1}\right)^{n}> \left(1+\frac{1}{n}\right)^{n+1}=f(n)$$ and we may conclude that $n\to f(n)$ is strictly decreasing.

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We could also apply the GM-HM inequality:

$$ \left(1+\frac{1}{n}\right)^{n+1}=1·\prod_{1}^{n+1}\frac{n+1}{n}>\left(\frac{n+2}{1+\sum_1^{n+1}\frac{n}{n+1}}\right)^{n+2}=\left(1+\frac{1}{n+1}\right)^{n+2} $$

(Left side is strictly greater than right side since $1\ne1+\frac{1}{n}$, for every $n\in N^+$.)

Thus $$ f(n)> f(n+1). $$

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