Wsrtjtyk

I accidentally discovered this equality, in which I can prove numerically using python.

$$lim_{ktoinfty}{ sqrt[n]{ prod_{i=1}^{n}{(x_i+k)}} - k } = frac{sum_{i=1}^{n}{x_i}}{n}$$

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.

Assume that $k$ is greater than twice $|x_1|+ldots+|x_n|$ and also greater than twice $M=x_1^2+ldots+x_n^2$. Over the interval $left(-frac{1}{2},frac{1}{2}right)$ we have
$$ e^x = 1+x+C(x)x^2, qquad log(1+x)=x+D(x)x^2 $$
with $|C(x)|,|D(x)|leq 1$. It follows that

$$begin{eqnarray*}text{GM}(x_1+k,ldots,x_n+k)&=&kcdot text{GM}left(1+tfrac{x_1}{k},ldots,1+tfrac{x_n}{k}right)\&=&kexpleft[frac{1}{n}sum_{j=1}^{n}logleft(1+frac{x_j}{k}right)right]\&=&kexpleft[frac{1}{k}sum_{j=1}^{n}frac{x_j}{n}+Thetaleft(frac{M}{nk^2}right)right]\&=&kleft[1+frac{1}{k}sum_{j=1}^{n}frac{x_j}{n}+Thetaleft(frac{M}{nk^2}right)right]end{eqnarray*}$$
and the claim is proved. It looks very reasonable also without a formal proof: the magnitude of the difference between the arithmetic mean and the geometric mean is controlled by $frac{text{Var}(x_1,ldots,x_n)}{text{AM}(x_1,ldots,x_n)}$. A translation towards the right leaves the variance unchanged and increases the mean.

You could even get an interesting asymptotics considering $$y_n=sqrt[n]{ prod_{i=1}^{n}{(x_i+k)}}$$ $$ log(y_n)=frac 1n sum_{i=1}^{n}logleft({(x_i+k)}right)=frac 1n sum_{i=1}^{n}left(log(k)+logleft(1+frac {x_i}kright)right)$$ Assuming that $x_1<x_2<cdots < x_n ll k$, use the Taylor expansion of $logleft(1+frac {x_i}kright)$. Then continuing with Taylor series
$$y_n=e^{log(y_n)}implies
y_n=k+frac{sum_{i=1}^{n} x_i}{n}+frac{left(sum_{i=1}^{n} x_iright)^2-nsum_{i=1}^{n} x^2_i}{2n^2 k}+Oleft(frac{1}{k^2}right)$$ making
$$ sqrt[n]{ prod_{i=1}^{n}{(x_i+k)}} - k =frac{sum_{i=1}^{n} x_i}{n}+frac{left(sum_{i=1}^{n} x_iright)^2-nsum_{i=1}^{n} x^2_i}{2n^2 k}+Oleft(frac{1}{k^2}right)$$

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 $approx 12.8639$

Your expression under limit can be written as $k(a-b) $ where $$a=sqrt[n]{prod_{i=1}^{n}left(1+frac{x_i}{k}right)}, b=1$$ and further $a-b=(a^n-b^n) / c$ where $a, b$ tend to $1$ and $$c=a^{n-1}+a^{n-2}b+dots+ab^{n-2}+b^{n-1}$$ tends to $n$ as $ktoinfty$. And $$a^n-b^n=prod_{i=1}^{n}left(1+frac{x_i}{k}right)-1=frac{1}{k}sum_{i=1}^{n}x_i+o(1/k)$$ and thus $$k(a-b) =frac{1}{n}sum_{i=1}^{n}x_i+o(1)$$ and the desired limit as $ktoinfty$ is $dfrac{1}{n}sumlimits_{i=1}^{n}x_i$.

You can also give an argument using L'Hopital's rule:
begin{align*}
lim_{k to infty} sqrt[n]{prod_{i = 1}^n (x_i + k)} - k &= lim_{k to infty} frac{sqrt[n]{prod_{i = 1}^n left(frac{x_i}{k} + 1right)} - 1}{frac{1}{k}} \
&overset{(H)}{=} lim_{k to infty} frac{frac{1}{n} cdot frac{sum_{i = 1}^n frac{-x_i}{k^2} cdot prod_{j ne i} left(frac{x_i}{k} + 1right)}{left(prod_{i = 1}^n left(frac{x_i}{k} + 1right)right)^{1 - 1/n}}}{frac{-1}{k^2}} \
&= lim_{k to infty}frac{1}{n} cdot frac{sum_{i = 1}^n x_i cdot prod_{j ne i} left(frac{x_i}{k} + 1right)}{left(prod_{i = 1}^n left(frac{x_i}{k} + 1right)right)^{1 - 1/n}} \
&= frac{sum_{i = 1}^n x_i}{n}
end{align*}

$$
begin{align}
lim_{ktoinfty}left[left(prod_{i=1}^n(x_i+k)right)^{1/n}-kright]
&=lim_{ktoinfty}kleft[prod_{i=1}^nleft(1+frac{x_i}kright)^{1/n}-1right]\
&=lim_{ktoinfty}frac{prodlimits_{i=1}^nleft(1+frac{x_i}kright)^{1/n}-1}{logleft(prodlimits_{i=1}^nleft(1+frac{x_i}kright)^{1/n}right)}cdotlim_{ktoinfty}frac knsum_{i=1}^nlogleft(1+frac{x_i}kright)\
&=lim_{xto1}frac{x-1}{log(x)}cdotfrac1nsum_{i=1}^nlim_{ktoinfty}klogleft(1+frac{x_i}kright)\[9pt]
&=frac1nsum_{i=1}^nx_i
end{align}
$$