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I want to prove that for positive integer $n$, $$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n}$$ but I am stuck on how to proceed. Can someone help me?

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    $\begingroup$ Bernoulli's Inequality gives the inequality immediately. Note $$\left(1-\frac1{n^2}\right)^n\ge 1-n\times \frac1{n^2}=1-\frac1n$$ $\endgroup$ – Mark Viola Feb 20 at 6:18
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Hint: \begin{eqnarray*} \left( 1- \frac{1}{n^2} \right)^n = 1+ n \left( - \frac{1}{n^2} \right) + \binom{n}{2} \left( - \frac{1}{n^2} \right)^2+ \cdots \end{eqnarray*}

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    $\begingroup$ What about the negative terms of the right hand side? For example $\binom{n}{3}\left(-\frac1{n^2}\right)^3$ is $<0$. Probably using the Bernoulli inequality $(1+x)^n\geq 1+n x$ when $x>-1$ would be helpfull. $\endgroup$ – Jens Schwaiger Feb 20 at 4:20
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    $\begingroup$ The alternating series theorem applies here. The error has the sign of the first neglected term, which is positive, so the right side is greater than $1-\frac 1n$ $\endgroup$ – Ross Millikan Feb 20 at 4:31
  • $\begingroup$ Thank you so much! $\endgroup$ – Xiaohuolong Feb 20 at 4:56
  • $\begingroup$ @ Ross Millikan: Is the sequence of absolute values of the terms in question monotonically decreasing? $\endgroup$ – Jens Schwaiger Feb 20 at 6:47
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Note that, for $n \gt 1$,

$$\frac{1}{1 - \frac{1}{n}} = \frac{n}{n - 1} = 1 + \frac{1}{n - 1} \tag{1}\label{eq1A}$$

Using \eqref{eq1A}, for $n \gt 1$, multiplying both sides of your proposed inequality by $\left(\frac{1}{1 - \frac{1}{n}}\right)^n$ gives

$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n} \iff \left(1+\frac{1}{n}\right)^n \ge \left(1+\frac{1}{n-1}\right)^{n-1} \tag{2}\label{eq2A}$$

For $x \gt 1$, consider

$$f(x) = \left(1 + \frac{1}{x}\right)^x \implies \log(f(x)) = x\log\left(1 + \frac{1}{x}\right) \tag{3}\label{eq3A}$$

You could just use that $f(x)$ is an increasing function or, if you wish to prove it, you could take the derivative of both sides to get

$$\begin{equation}\begin{aligned} \frac{f'(x)}{f(x)} & = \log\left(1 + \frac{1}{x}\right) + \frac{x}{1 + \frac{1}{x}}\left(-\frac{1}{x^2}\right) \\ & = \log\left(1 + \frac{1}{x}\right) - \frac{\frac{1}{x}}{1 + \frac{1}{x}} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

The Inequalities section of Wikipedia's "List of logarithmic identities" article gives that, for $-1 \lt y$, you have

$$\frac{y}{1+y} \le \ln(1 + y) \tag{5}\label{eq5A}$$

Using $y = \frac{1}{x}$ shows the RHS of \eqref{eq4A} is non-negative. Since $f(x) \gt 0$, this means $f'(x) \ge 0$, i.e., $f(x)$ is an increasing function for $x \gt 1$. This shows that \eqref{eq2A} is also true for all $n \gt 1$. Of course, the inequality is also true trivially for $n = 1$, showing it's always true for positive integers.

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If you want a neat proof, use the fact that $\left(1-\frac{1}{n}\right)^n$ is an increasing sequence, and that $$\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right) = \left(1-\frac{1}{n^2}\right).$$ We have that for some $n$: $$\left[\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^n\right]^n = \left(1-\frac{1}{n^2}\right)^{n^2} > \left(1-\frac{1}{n}\right)^n,$$ so this must be true for the $n$-th root as well, which is what you require.

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    $\begingroup$ Wow this is very elegant proof! $\endgroup$ – Xiaohuolong Feb 20 at 4:56
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Consider $$a_n=\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n=\left(1-\frac{1}{n^2}\right)^n\implies \log(a_n)=n \log\left(1-\frac{1}{n^2}\right)$$ So, by Taylor $$\log(a_n)=n\left(-\frac{1}{n^2}-\frac{1}{2 n^4}-\frac{1}{3 n^6}+O\left(\frac{1}{n^8}\right)\right)=-\frac{1}{n}-\frac{1}{2 n^3}-\frac{1}{3 n^5}+O\left(\frac{1}{n^7}\right)$$ $$a_n=e^{\log(a_n)}=1-\frac{1}{n}+\frac{1}{2 n^2}-\frac{2}{3 n^3}+\frac{13}{24 n^4}++O\left(\frac{1}{n^5}\right)$$ and, as Ross Millikan commented, this is an alternating series and then ...

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