(1) For $ a>0,\ b\geq 0$, Compare

$ \int_b^{b\plus{}1} \frac{dx}{\sqrt{x\plus{}a}},\ \frac{1}{\sqrt{a\plus{}b}},\ \frac{1}{\sqrt{a\plus{}b\plus{}1}}$.

(2) Find $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{1}{\sqrt{n^2\plus{}k}}$.
### Part (1)
We need to compare the following three expressions for \( a > 0 \) and \( b \geq 0 \):
1. \( \int_b^{b+1} \frac{dx}{\sqrt{x+a}} \)
2. \( \frac{1}{\sqrt{a+b}} \)
3. \( \frac{1}{\sqrt{a+b+1}} \)

#### Step 1: Evaluate the integral
First, we evaluate the integral \( \int_b^{b+1} \frac{dx}{\sqrt{x+a}} \).

Let \( u = x + a \). Then \( du = dx \) and the limits of integration change as follows:
- When \( x = b \), \( u = b + a \)
- When \( x = b+1 \), \( u = b + 1 + a \)

Thus, the integral becomes:
\[
\int_b^{b+1} \frac{dx}{\sqrt{x+a}} = \int_{b+a}^{b+1+a} \frac{du}{\sqrt{u}} = \int_{b+a}^{b+1+a} u^{-\frac{1}{2}} \, du
\]

Using the power rule for integration, we get:
\[
\int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} + C
\]

Evaluating this from \( b+a \) to \( b+1+a \), we have:
\[
2 \left[ u^{\frac{1}{2}} \right]_{b+a}^{b+1+a} = 2 \left( (b+1+a)^{\frac{1}{2}} - (b+a)^{\frac{1}{2}} \right)
\]

Thus, the integral is:
\[
\int_b^{b+1} \frac{dx}{\sqrt{x+a}} = 2 \left( \sqrt{b+1+a} - \sqrt{b+a} \right)
\]

#### Step 2: Compare the integral with the given terms
We need to compare \( 2 \left( \sqrt{b+1+a} - \sqrt{b+a} \right) \) with \( \frac{1}{\sqrt{a+b}} \) and \( \frac{1}{\sqrt{a+b+1}} \).

1. **Comparison with \( \frac{1}{\sqrt{a+b}} \)**:
   \[
   2 \left( \sqrt{b+1+a} - \sqrt{b+a} \right) \quad \text{vs} \quad \frac{1}{\sqrt{a+b}}
   \]

   Since \( \sqrt{b+1+a} > \sqrt{b+a} \), we have:
   \[
   2 \left( \sqrt{b+1+a} - \sqrt{b+a} \right) > 0
   \]

   To compare with \( \frac{1}{\sqrt{a+b}} \), note that for large \( a \) and \( b \), the difference \( \sqrt{b+1+a} - \sqrt{b+a} \) becomes small, but the factor of 2 ensures that the integral is larger than \( \frac{1}{\sqrt{a+b}} \).

2. **Comparison with \( \frac{1}{\sqrt{a+b+1}} \)**:
   \[
   2 \left( \sqrt{b+1+a} - \sqrt{b+a} \right) \quad \text{vs} \quad \frac{1}{\sqrt{a+b+1}}
   \]

   Similarly, for large \( a \) and \( b \), the term \( \sqrt{b+1+a} - \sqrt{b+a} \) is small, but the factor of 2 ensures that the integral is larger than \( \frac{1}{\sqrt{a+b+1}} \).

Thus, we conclude:
\[
\int_b^{b+1} \frac{dx}{\sqrt{x+a}} > \frac{1}{\sqrt{a+b}} \quad \text{and} \quad \int_b^{b+1} \frac{dx}{\sqrt{x+a}} > \frac{1}{\sqrt{a+b+1}}
\]

### Part (2)
We need to find the limit:
\[
\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{n^2 + k}}
\]

#### Step 1: Approximate the sum
For large \( n \), \( n^2 + k \approx n^2 \). Thus, we can approximate:
\[
\frac{1}{\sqrt{n^2 + k}} \approx \frac{1}{\sqrt{n^2}} = \frac{1}{n}
\]

The sum becomes:
\[
\sum_{k=1}^n \frac{1}{\sqrt{n^2 + k}} \approx \sum_{k=1}^n \frac{1}{n} = \frac{n}{n} = 1
\]

#### Step 2: Evaluate the limit
To rigorously evaluate the limit, we can use the Squeeze Theorem. Note that:
\[
\frac{1}{\sqrt{n^2 + n}} \leq \frac{1}{\sqrt{n^2 + k}} \leq \frac{1}{\sqrt{n^2 + 1}}
\]

Summing from \( k = 1 \) to \( n \), we get:
\[
\sum_{k=1}^n \frac{1}{\sqrt{n^2 + n}} \leq \sum_{k=1}^n \frac{1}{\sqrt{n^2 + k}} \leq \sum_{k=1}^n \frac{1}{\sqrt{n^2 + 1}}
\]

Each term in the sums on the left and right is constant:
\[
\frac{n}{\sqrt{n^2 + n}} \leq \sum_{k=1}^n \frac{1}{\sqrt{n^2 + k}} \leq \frac{n}{\sqrt{n^2 + 1}}
\]

As \( n \to \infty \):
\[
\frac{n}{\sqrt{n^2 + n}} \to 1 \quad \text{and} \quad \frac{n}{\sqrt{n^2 + 1}} \to 1
\]

By the Squeeze Theorem:
\[
\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{n^2 + k}} = 1
\]

The final answer is \( \boxed{ 1 } \)