11. (20 points) Given that for any positive integer $n$ we have $\left(1+\frac{1}{n}\right)^{n-a} \geqslant \mathrm{e}$.
Find the range of values for the real number $a$.


11. Taking the natural logarithm of both sides of the given inequality, we get
$$
\begin{array}{l}
(n-a) \ln \left(1+\frac{1}{n}\right) \geqslant 1 \\
\Rightarrow\left(1-\frac{a}{n}\right) \ln \left(1+\frac{1}{n}\right) \geqslant \frac{1}{n} \\
\Rightarrow\left(1-\frac{a}{n}\right) \ln \left(1+\frac{1}{n}\right)-\frac{1}{n} \geqslant 0 .
\end{array}
$$
$$
\text { Let } f(x)=(1-a x) \ln (1+x)-x(0<x \leqslant 1) \text {. }
$$

For inequality (1) to hold, it is sufficient that for any $0<x \leqslant 1$,
$$
f(x) \geqslant 0 \text {. }
$$

Notice that,
$$
\begin{array}{l}
f^{\prime}(x)=-a \ln (1+x)+\frac{1-a x}{1+x}-1 \\
=-a \ln (1+x)+\frac{1+a}{1+x}-(a+1), \\
f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^{\prime} \\
=\frac{-a}{1+x}-\frac{1+a}{(1+x)^{2}} \\
=\frac{-a x-2 a-1}{(1+x)^{2}} .
\end{array}
$$
(1) If $a \leqslant-\frac{1}{2}$, then for $0<x \leqslant 1$, we have
$$
f^{\prime \prime}(x)=\frac{-a x-(2 a+1)}{(1+x)^{2}} \geqslant 0 \text {. }
$$

Thus, $f^{\prime}(x)$ is monotonically increasing.
Therefore, $f^{\prime}(x) \geqslant f^{\prime}(0)=0$.
Hence, $f(x)$ is monotonically increasing, i.e., $f(x) \geqslant f(0)=0$, and inequality (2) clearly holds.
(2) If $a \geqslant 0$, then for $0 \leqslant x \leqslant 1$, we have $f^{\prime \prime}(x)<0$.
Thus, $f^{\prime}(x)$ is monotonically decreasing.
Therefore, $f^{\prime}(x) \leqslant f^{\prime}(0)=0$.
Hence, $f(x)$ is monotonically decreasing, i.e., $f(x) \leqslant f(0)=0$, which does not satisfy inequality (2).
(3) If $-\frac{1}{2}<a<0$, let $b=\min \left\{1,-\frac{2 a+1}{a}\right\}$.
When $0<x<b$, $f^{\prime \prime}(x) \leqslant 0$, thus, $f^{\prime}(x)$ is monotonically decreasing.
Therefore, $f^{\prime}(x) \leqslant f^{\prime}(0)=0$.
Hence, $f(x)$ is monotonically decreasing, i.e., $f(x) \leqslant f(0)=0$, which does not satisfy inequality (2).
In summary, $a \leqslant-\frac{1}{2}$.