Simplifying a sigma expression with square roots in the denominator? Let 
$$A = \frac{1}{\sqrt{n}\sqrt{n + 1}} + \frac{1}{\sqrt{n}\sqrt{n + 2}}+...+\frac{1}{\sqrt{n}\sqrt{n + n}}$$
where $n$ is any real number such that,
$$A = \sum_{i = 1}^{n}\frac{1}{\sqrt{n}\sqrt{n+i}}$$ 
I'm looking to ditch the sigma and simplify the expression for $A$ into a single formula with which I can plug any $n$ into. The reason for this is so I can solve the limit of a Riemann Sum and find $\triangle{x}$, where $\triangle{x} = \frac{b - a}{n}$, and $a$ and $b$ represent the start and end of the definite integral, respectively.  
What are some good methods for approaching this kind of problem? The main focus on this is to get rid of the sigma and find $\triangle{x}$.
I appreciate any advice.
 A: It's generally very hard, if not impossible, to find closed expressions for sums like this. 
If your eventual goal is to take the limit and evaluate the integral, this isn't the way to do it. You show that the limit of the sum is equal to some integral, then you evaluate the integral using techniques of integral calculus. 
Here's what you would do if you wanted to evaluate the integral instead:
You notice that
$$\frac{1}{\sqrt{n}\sqrt{n+i}} = \frac{1/i}{\sqrt{(n/i)}\sqrt{(n/i) + 1}} = \frac{1}{i} f\left(\frac{n}{i}\right),$$
where $f(x) = \frac{1}{\sqrt{x}\sqrt{x+1}} = \frac{1}{\sqrt{x^2+x}}.$ 
Then the limit of the sum you mentioned is equal to the integral of $f(x)$ from $0$ to $1$:
$$\lim_{n\rightarrow \infty}\sum_{i=1}^n \frac{1}{\sqrt{n}\sqrt{n+i}}=\int_{0}^1 \frac{1}{\sqrt{x^2+x}} dx $$
Now you can evaluate the integral using integration techniques. For example writing $x^2+x = (x^2+x+\frac{1}{4})-\frac{1}{4}=(x+\frac{1}{2})^2 - \frac{1}{4},$ one can get the idea to set $y=2(x+\frac{1}{2}),$ then we have $dy = 2dx$ and so
$$\frac{1}{\sqrt{x^2+x}} dx = \frac{1}{\sqrt{\frac{y^2}{4}-\frac{1}{4}}} \frac{dy}{2}= \frac{1}{\sqrt{y^2-1}} dy,$$
so the entire integral becomes
$$\int_1^3 \frac{1}{\sqrt{y^2-1}} dy,$$
which one can proceed to evalute in various ways.
Let me mention that this kind of exercise in first year calculus courses is only there to test your understanding of the definition of the integral as Riemann sums. One almost never needs to evaluate such sums. It's often the integral itself that's more important, and one evaluates those using integration techniques, almost never using the definition of the integral.
You mentioned the expression $\Delta x =\frac{b-a}{n}$ in your question. In exercises like this almost always $b=1$ and $a=0$, as it was in this question.
A: about blank!
This problem hides a tiny, little manipulation of the sum:
$$ \sum_{i=1}^{n} \frac{1}{\sqrt{n} \sqrt{n+i}} = \frac{1}{n} \sum_{i=1}^{n} \frac{1}{ \sqrt{1+\frac{i}{n} } } $$ 
From here, we can guess $\Delta x$ and the integrand. Hence, we can find our value of the definite integral when the limit of $n$ is taken to $\infty$.
