You are correct, the $\frac{1}{n}$ term makes the problem more complicated. However, it does not contribute to the limit in the sense that removing it will not change the limit. The following solution uses that.
Let $$S(p,q)=\lim_{n\to \infty} \sum_{k=0}^n \frac{1}{qn+pk+1},\qquad T(p,q)=\lim_{n\to \infty}\sum_{k=0}^n \frac{1}{qn+pk}$$
Claim 1: $T(p,q)=\int_0^1\frac{dx}{q+p x}$ for any $p,q>0$.
Claim 1 can be proved using Riemann sums
$$T(p,q)=\lim_{n\to \infty}\frac{1}{n} \sum_{k=0}^n \frac{1}{q+p\frac{k}{n}}.$$
Claim 2: $T(p,q)=S(p,q)$ for any $p,q>0$. To see this, notice that
$$\frac{1}{qn+pk}-\frac{1}{qn+pk+1} = \frac{1}{(q n+p k) (qn+pk+1)}\le \frac{1}{q^2n^2}$$
So,
$$\left|\sum_{k=0}^n \frac{1}{qn+pk+1}-\sum_{k=0}^n \frac{1}{qn+pk}\right|\le \frac{1}{q^2 n}$$
By taking $n\to \infty$, we can see that claim 2 holds.
That's not exactly what the definition is. The limit should be
$$\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right)$$
i.e., you are summing over $n$ elements. Sure, each of the summands becomes smaller and smaller, but the number of the summands increases at the same rate, so the sum will not always be zero.
In fact, take a look at what happens when $f(x)=1$ for all $x$. In that case,
$$\begin{align}\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right)&=\lim_{n\to\infty}\sum_{i=1}^n 1\cdot\left(\frac{b-a}{n}\right)\\&=\lim_{n\to\infty}\left((b-a)\cdot\sum_{i=1}^n\frac{1}{n}\right)\\&=\lim_{n\to\infty}((b-a)\cdot 1) = \lim_{n\to\infty}(b-a)=b-a\neq 0\end{align}$$
Also, a further word of warning not entirely on topic, but related to your question:
If the Riemann integral of a function over an interval $[a,b]$ exists, then the integral is equal to $$\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right).$$
However, the existance of the limit above does not, in itself, guarantee that a function is Riemann integrable. There exist functions for which
$$\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\left(\frac{b-a}{n}\right)$$
exists, but $\int_a^b f(x)dx$ does not exist.
Best Answer
Note: $$\sum\frac{k\cdot \sqrt{n+k}}{n^{5/2}}=\sum\frac{1}{n} \cdot\frac{k}{n} \cdot\sqrt{1+\frac{k}{n}}$$
Now this is of the form $\sum\frac{1}{n}\cdot f(\frac{k}{n}) = \int_{0}^{1}f(x) \ dx$