Prove that $\lim_{k\to\infty}{ \sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}} - k } = \frac{\sum_{i=1}^{n}{x_i}}{n}$ I accidentally discovered this equality, in which I can prove numerically using python. 
$$\lim_{k\to\infty}{ \sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}} - k } = \frac{\sum_{i=1}^{n}{x_i}}{n}$$
But I need to prove this equality in an algebraic way and I could not get anywhere. The right-hand side is nothing more than an arithmetic mean, while the left-hand side is a modification of the geometric mean.
I hope someone can help me at this point.

You could even get an interesting asymptotics considering $$y_n=\sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}}$$ $$ \log(y_n)=\frac 1n \sum_{i=1}^{n}\log\left({(x_i+k)}\right)=\frac 1n \sum_{i=1}^{n}\left(\log(k)+\log\left(1+\frac {x_i}k\right)\right)$$ Assuming that $x_1<x_2<\cdots < x_n \ll k$,  use the Taylor expansion of $\log\left(1+\frac {x_i}k\right)$. Then continuing with Taylor series
$$y_n=e^{\log(y_n)}\implies
y_n=k+\frac{\sum_{i=1}^{n} x_i}{n}+\frac{\left(\sum_{i=1}^{n} x_i\right)^2-n\sum_{i=1}^{n} x^2_i}{2n^2 k}+O\left(\frac{1}{k^2}\right)$$ making
$$ \sqrt[n]{ \prod_{i=1}^{n}{(x_i+k)}} - k =\frac{\sum_{i=1}^{n} x_i}{n}+\frac{\left(\sum_{i=1}^{n} x_i\right)^2-n\sum_{i=1}^{n} x^2_i}{2n^2 k}+O\left(\frac{1}{k^2}\right)$$ 
For example, using $n=10$, $k=1000$ and $x_i=p_i$, the approximation would give $\frac{2572671}{200000}\approx 12.8634$ while the exact calculation would lead to $\approx 12.8639$
