Limit value of an expectation involving the $k$-th powers of uniform random variables. Let $X_1,X_2,X_3.\dots$ be iid Uniform $(0,1)$ and let $N_k$ be the minimum $n$ such that $$X_1^k+X_2^k+\cdots +X_n^k\le 1<X_1^k+X_2^k+\cdots +X_{n+1}^k, \quad  k\in\mathbb{N}$$ 
How to find $\displaystyle \lim_{k\to\infty}\frac{E(N_k)}{k}$? 
 A: If $X_i$ is uniformly distributed over $(0,1)$, then
$$ \mathbb{P}[X_i^k\leq t] = t^{\frac{1}{k}}$$
hence the pdf of $X_i^k$ is supported on $(0,1)$ and given by $\frac{1}{k}\,t^{\frac{1}{k}-1}$. Moreover,
$$\begin{eqnarray*}\mathbb{P}[N_k\geq n]=\mathbb{P}[X_1^k+\ldots+X_n^k<1]&=&\int_{(0,1)^n}\mathbb{1}_{x_1^k+\ldots+x_n^k<1}\,d\mu\\&=&\frac{1}{k^n}\int_{(0,1)^n}\left(z_1\cdot\ldots\cdot z_n\right)^{\frac{1}{k}-1}\cdot\mathbb{1}_{z_1+\ldots+z_n<1}\,d\mu\\&=&\frac{\Gamma\left(\frac{1}{k}\right)^{n}}{k^n\cdot \Gamma\left(1+\frac{n}{k}\right)}\end{eqnarray*}$$
where the last step follows from the properties of the Dirichlet distribution.
It gives:
$$\mathbb{E}[N_k]=\sum_{n\geq 1}\frac{\Gamma\left(1+\frac{1}{k}\right)^n}{\Gamma\left(1+\frac{n}{k}\right)}$$
but since the limit for $k\to +\infty$ of $\Gamma\left(1+\frac{1}{k}\right)^k$ equals $e^{-\gamma}$, by a Riemann sum argument:

$$ \lim_{k\to +\infty}\frac{\mathbb{E}[N_k]}{k}=\int_{0}^{+\infty}\frac{e^{-\gamma x}}{\Gamma(1+x)}\,dx = \color{red}{1.24941661444\ldots}.$$

