Example 8 Let $f(x)=\lg \frac{1+2^{x}+3^{x}+\cdots+(n-1)^{x}+a \cdot n^{x}}{n}$, where $a$ is a real number, and $n$ is any given natural number with $n \geqslant 2$. If $f(x)$ is meaningful for $x \in(-\infty, 1]$, find the range of values for $a$.

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From the problem, we know that when $x \in (-\infty, 1]$, $n \geqslant 2$, $1+2^{x}+3^{x}+\cdots+(n-1)^{x}+a \cdot n^{x}>0$ always holds. Therefore, $a>-\left[\left(\frac{1}{n}\right)^{x}+\left(\frac{2}{n}\right)^{x}+\cdots+\left(\frac{n-1}{n}\right)^{x}\right]$ always holds, $x \in (-\infty, 1]$. Since the function $-\left(\frac{k}{n}\right)^{x} (k=1,2, \cdots, n-1)$ is increasing on $(-\infty, 1]$, $-\left[\left(\frac{1}{n}\right)^{x}+\left(\frac{2}{n}\right)^{x}+\cdots+\left(\frac{n-1}{n}\right)^{x}\right]$ is also increasing on $(-\infty, 1]$, and thus it achieves its maximum value at $x=1$, which is $-\left(\frac{1}{n}+\frac{2}{n}+\cdots+\frac{n-1}{n}\right)=\frac{\frac{1}{2} n(n-1)}{n}=-\frac{1}{2}(n-1)$. Therefore, $a>-\frac{1}{2}(n-1)$, which means the range of $a$ is $\left\{a \left\lvert\, a>-\frac{1}{2}(n-1)\right.\right\}$.