For each $n\in\mathbb N$, the function $f_n$ is defined on real numbers $x\ge n$ by
$$f_n(x)=\sqrt{x-n}+\sqrt{x-n+1}+\ldots+\sqrt{x+n}-(2n+1)\sqrt x.$$(a) If $n$ is fixed, prove that $\lim_{x\to+\infty}f_n(x)=0$.
(b) Find the limit of $f_n(n)$ as $n\to+\infty$.
(a) To prove that $\lim_{x \to +\infty} f_n(x) = 0$ for a fixed $n$, we start by rewriting the function $f_n(x)$:

\[ f_n(x) = \sum_{k=-n}^{n} \sqrt{x+k} - (2n+1)\sqrt{x} \]

We can rewrite each term inside the summation as follows:

\[ \sqrt{x+k} = \sqrt{x} \sqrt{1 + \frac{k}{x}} \]

Using the first-order Taylor expansion for the square root function, $\sqrt{1 + y} \approx 1 + \frac{y}{2}$ for small $y$, we get:

\[ \sqrt{x+k} \approx \sqrt{x} \left(1 + \frac{k}{2x}\right) = \sqrt{x} + \frac{k}{2\sqrt{x}} \]

Substituting this approximation into the summation:

\[ \sum_{k=-n}^{n} \sqrt{x+k} \approx \sum_{k=-n}^{n} \left( \sqrt{x} + \frac{k}{2\sqrt{x}} \right) \]

This can be split into two sums:

\[ \sum_{k=-n}^{n} \sqrt{x+k} \approx \sum_{k=-n}^{n} \sqrt{x} + \sum_{k=-n}^{n} \frac{k}{2\sqrt{x}} \]

The first sum is:

\[ \sum_{k=-n}^{n} \sqrt{x} = (2n+1)\sqrt{x} \]

The second sum is:

\[ \sum_{k=-n}^{n} \frac{k}{2\sqrt{x}} = \frac{1}{2\sqrt{x}} \sum_{k=-n}^{n} k \]

Since the sum of integers from $-n$ to $n$ is zero:

\[ \sum_{k=-n}^{n} k = 0 \]

Thus, the second sum is zero:

\[ \sum_{k=-n}^{n} \frac{k}{2\sqrt{x}} = 0 \]

Combining these results, we get:

\[ \sum_{k=-n}^{n} \sqrt{x+k} \approx (2n+1)\sqrt{x} \]

Therefore:

\[ f_n(x) = \sum_{k=-n}^{n} \sqrt{x+k} - (2n+1)\sqrt{x} \approx (2n+1)\sqrt{x} - (2n+1)\sqrt{x} = 0 \]

As $x \to +\infty$, the approximation becomes exact, so:

\[ \lim_{x \to +\infty} f_n(x) = 0 \]

(b) To find the limit of $f_n(n)$ as $n \to +\infty$, we start by evaluating $f_n(n)$:

\[ f_n(n) = \sum_{k=-n}^{n} \sqrt{n+k} - (2n+1)\sqrt{n} \]

For large $n$, the terms $\sqrt{n+k}$ can be approximated by $\sqrt{n}$ since $k$ ranges from $-n$ to $n$ and the relative difference becomes small. Thus:

\[ \sqrt{n+k} \approx \sqrt{n} \]

So:

\[ f_n(n) \approx \sum_{k=-n}^{n} \sqrt{n} - (2n+1)\sqrt{n} = (2n+1)\sqrt{n} - (2n+1)\sqrt{n} = 0 \]

Therefore, as $n \to +\infty$:

\[ \lim_{n \to +\infty} f_n(n) = 0 \]

The final answer is $\boxed{0}$