TITLE: What could be this number?
QUESTION [16 upvotes]: In a much larger problem, I need to solve for $x_n$ the equation $$\sum_{i=0}^n (n+i)^{x_n}=(2n+1)^{x_n}$$ where $n$ can be large.
From a numerical point of view, it does not make any practical difficulty solving instead $$\log\left(\sum_{i=0}^n (n+i)^{x_n} \right)={x_n}\log(2n+1)$$ which is "almost" finding the intersection of two straight lines.
My first surprise, working with small values of $n$, was to notice that the solution is almost linearly dependent on $n$.
Working with large numbers,  what I noticed is the following
$$\left(
\begin{array}{cc}
 n & \frac {x_n} n \\
10^1 & 1.351611135 \\
10^2 & 1.382828492 \\
10^3 & 1.385947786 \\
10^4 & 1.386259704
\end{array}
\right)$$ and the ratio "seems" to converge to something.

So, my questions :

Could we prove the linear dependency of $x_n$ to $n$ ?
Could we prove that $\frac {x_n} n$ tends to a limit ?
If there is a limit, what could be this number ? In other words, would it be possible to establish the asymptotics of $x_n$ ?


Edit after answers
After the beautiful answers I received from Professor Vector and Raymond Manzoni (who made the problem more general), I solved, for $n=10^4$, the equation$$\sum_{i=0}^n (n+i)^{x_n}=(2n+a)^{x_n}$$ In the table below are reproduced the results for the computed ratio $\frac {x_n} n$ and the value computed for $y_a=-2\log(b)$ where $b$ is the positive solution of $1-b=b^a$.
$$\left(
\begin{array}{cc}
 a & \frac {x_n} n & y_a\\
 1 & 1.38626 &1.38629 \\
 2 & 0.96240 &0.96242 \\
 3 & 0.76448 &0.76449 \\
 4 & 0.64457 &0.64457 \\
 5 & 0.56240 &0.56240 \\
 6 & 0.50184 &0.50183 \\
 7 & 0.45496 &0.45495 \\
 8 & 0.41739 &0.41737 \\
 9 & 0.38646 &0.38644
\end{array}
\right)$$ 
Concerning the case where $a=1$, it is interesting to compare the results I gave with the asymptotics  Raymond Manzoni gave in a comment. $$\frac {x_n} n=2\log(2)-\frac{\log(2)}{2n}+O\left(\frac 1 {n^2} \right)$$
$$\left(
\begin{array}{cc}
 n & \frac {x_n} n &2\log(2)-\frac{\log(2)}{2n}\\
10^1 & 1.351611135 &1.351637002\\
10^2 & 1.382828492 &1.382828625\\
10^3 & 1.385947786 &1.385947788\\
10^4 & 1.386259704 &1.386259704
\end{array}
\right)$$
I do not think that I could add any comment !

REPLY [4 votes]: Using Bernoulli's Inequality,
$$
\begin{align}
\frac{\left(1-\frac{k+1}{2n+3}\right)^{n+\frac32}}{\left(1-\frac{k+1}{2n+1}\right)^{n+\frac12}}
&=\left(1-\frac{k+1}{2n+3}\right)\left(\frac{2n+2-k}{2n+3}\frac{2n+1}{2n-k}\right)^{n+\frac12}\\
&=\left(1-\frac{k+1}{2n+3}\right)\left(1+\frac{2k+2}{(2n+3)(2n-k)}\right)^{n+\frac12}\\[9pt]
&\ge\left(1-\frac{k+1}{2n+3}\right)\left(1+\frac{(2n+1)(k+1)}{(2n+3)(2n-k)}\right)\\[9pt]
&=1+\frac{2(k+1)^2}{(2n-k)(2n+3)^2}\\[12pt]
&\ge1
\end{align}
$$
That is, $\left(1-\frac{k+1}{2n+1}\right)^{n+\frac12}$ is increasing in $n$.
Therefore, by Monotone Convergence,
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=0}^n\left(\frac{n+k}{2n+1}\right)^{\left(n+\frac12\right)a}
&=\lim_{n\to\infty}\sum_{k=0}^n\left(\frac{2n-k}{2n+1}\right)^{\left(n+\frac12\right)a}\\
&=\lim_{n\to\infty}\sum_{k=0}^n\left(1-\frac{k+1}{2n+1}\right)^{\left(n+\frac12\right)a}\\
&=\sum_{k=0}^\infty e^{-(k+1)a/2}\\
&=\frac{e^{-a/2}}{1-e^{-a/2}}
\end{align}
$$
The $a$ that gives the limit equal to $1$ is
$$
a=2\log(2)
$$