Help me find the following limit : $\lim_{{n}\to{\infty}} (\frac{2^x+3^x+\cdots+n^x}{n-1})^\frac{1}{x} = ?$ I have no idea where to start.$$\begin{align}\lim_{{n}\to{\infty}} \left(\dfrac{2^x+3^x+\cdots+n^x}{n-1}\right)^{1/x} = ?, n >1\\\end{align}$$
 A: No need of Cesaro-Stoltz. Assuming $x\geq 1$, the function $n\to n^x$ is convex, hence:
$$\frac{2^x+3^x+\ldots+n^{x}}{n-1}\geq \left(\frac{2+3+\ldots+n}{n-1}\right)^x =\left(1+\frac{n}{2}\right)^x$$
by Jensen's inequality and the limit is $+\infty$.
A: Hint: Assume that $x > 0$, $\ln f(x) = \dfrac{1}{x}\cdot \ln\left(\dfrac{2^x+3^x+\cdots n^x}{n-1}\right)=\dfrac{1}{x}\cdot \left(\ln\left(\dfrac{2^x+3^x+\cdots n^x}{n^x}\right)+x\ln\left(\dfrac{n}{n-1}\right)\right)$.
Now we find the limit as $n \to \infty$ of the quotient: $q_n(x)=\dfrac{2^x+3^x+\cdots n^x}{n^x}$ by using Stolz-Cesaro's theorem: 
$\displaystyle \lim_{n\to\infty} q_n(x) = \displaystyle \lim_{n\to \infty} \dfrac{(n+1)^x}{(n+1)^x-n^x}$
A: Another approach:
$$\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}\le\left(\frac{n-1}{(n-1)2^x}\right)^{1/x}=\frac12\implies \sum_{n=1}^\infty\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}$$
converges, and then
$$\left(\frac{n-1}{2^x+3^x+\ldots+n^x}\right)^{1/x}\xrightarrow[n\to\infty]{}0\implies\left(\frac{2^x+3^x+\ldots+n^x}{n-1}\right)^{1/x}\xrightarrow[n\to\infty]{}\infty$$
since everything positive here.
