Finding Value Of Infinite Sum 
Let $x$ be a real number. Find $$\lim_{n\to\infty}\dfrac{\sum_{k=1}^{n}\sqrt[x+1]{1^{x}+2^{x}+\cdots +k^{x}}}{n^{2}}$$

It seems different from calculating it by transforming definite integral.
 A: The key observation is the following
Lemma: For each positive integer $x$,
$$
L:=\lim_{k\to\infty}\frac{1}{k}\sqrt[x+1]{1+2^x+\ldots+k^x}
$$
exists and is equal to $L=\frac{1}{\sqrt[x+1]{x+1}}$.
Proof: This follows immediately from Faulhaber's formula.$\quad\square$
In particular, there exists a sequence $(c_k)$ converging to zero such that
$$
\sqrt[x+1]{1+2^x+\ldots+k^x}=k(L+c_k).
$$
This implies that the limit in the OP's question satisfies
$$
\Lambda:=\lim_{n\to\infty}n^{-2}\sum_{k=1}^n{\sqrt[x+1]{1+2^x+\ldots+k^x}} = \lim_{n\to\infty}n^{-2}\sum_{k=1}^nk\left(L+c_k\right).
$$
For each $\varepsilon>0$ we can choose $k_0$ such that $|c_k|\leqslant\varepsilon$ for all $k>k_0$; we can then write
$$
\Lambda=\lim_{n\to\infty}n^{-2}\sum_{k=1}^{k_0}{k(L+c_k)}+\lim_{n\to\infty}n^{-2}\sum_{k=k_0+1}^n{k(L+c_k)}.
$$
The first term is clearly zero. The second term is bounded by
$$
(L\pm\varepsilon)\lim_{n\to\infty}\frac{(n+k_0+1)(n-k_0)}{2n^2}=\frac{L\pm\varepsilon}{2}.
$$
and is thus equal to $L/2$.
Thus,
$$
\Lambda = 0+\frac{L}{2} = \frac{1}{2}\frac{1}{\sqrt[x+1]{x+1}}.
$$
A similar argument should apply to non-integer $x$.

I actually believe that the following stronger result is true, but didn't try to prove it.
Conjecture: For each positive $x$,
$$
\lim_{k\to\infty}\left[\sqrt[x+1]{1+2^x+\ldots+k^x}-\frac{k}{\sqrt[x+1]{x+1}}\right]=\frac{1}{2}\frac{1}{\sqrt[x+1]{x+1}}.
$$
And apparently, the pattern can be continued in the form
$$
\lim_{k\to\infty}k\left[\sqrt[x+1]{1+2^x+\ldots+k^x}-\left(k+\frac{1}{2}\frac{1}{\sqrt[x+1]{x+1}}\right)\right] = -\frac{1}{24}\frac{x}{\sqrt[x+1]{x+1}},
$$
$$
\lim_{k\to\infty}k\left\{k\sqrt[x+1]{1+2^x+\ldots+k^x}-\left[k\left(k+\frac{1}{2}\right)-\frac{x}{24}\right]\frac{1}{(x+1)^{\frac{1}{x+1}}}\right\}=\frac{1}{48}\frac{x}{\sqrt[x+1]{x+1}},
$$
and so on.
