Limit $\lim_{x\to \infty}\left(\frac{1^x+2^x+3^x+...n^x}{n}\right)^{\dfrac{a}{x}}$

Question

Find $$\lim_{x\to \infty}\left(\frac{1^x+2^x+3^x+...n^x}{n}\right)^{\dfrac{a}{x}}$$

I do not know where to start or where to end. I have never seen something like this before. Any help or hint is much appreciated!

Answer 1

Let $f(x)$ denote the argument of the limit.

The following inequalities trivially hold: $$ \left(\frac{n^x}{n}\right)^{a/x} \leq f(x) \leq \left(\frac{n \cdot n^x}{n}\right)^{a/x}. $$ Hence $$ (n^a)^{(x-1)/x} \leq f(x) \leq n^a, $$ so that $\lim_{x\to \infty} f(x) = n^a$ by comparison.

Answer 2

The first inequality holds since

$f(x) = (\frac{1^n+2^n+...+n^x}{n})^{a/x}$ $= (\frac{1^n+2^n+...(n-1)^x}{n} + \frac{n^x}{n})^{a/x} \geq (\frac{n^x}{n})^{a/x}$

Answer 3

Let us set $$ f(x)=\left(\frac{1+2^x+\ldots+n^x}{n}\right)^{a/x}\,. $$ Taking the logarithm, $$ \log f(x)= \frac{a}{x}\log\left(\frac{1+2^x+\ldots+n^x}{n}\right) $$ which, as $x\to\infty$ is of the type $\infty/\infty$. Then, applying l'Hospital's rule, we have $$ a\,\frac{2^x\log2+3^x\log3+\ldots+n^x\log n}{1+2^x+\ldots+n^x}=a \frac{(2/n)^x\log2+(3/n)^x\log3+\ldots\log n}{(1/n)^x+(2/n)^x+\ldots+1} $$ whose limit as $x\to\infty$ is $a\log n$. Therefore $f(x)$ tends to $n^a$ as $x\to\infty$.