Which one is larger? 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?
 A: 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*}
A: 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.
A: 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.
A: 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 ...
