I want to illustrate my problem with help of an example, consider

$$ \lim_{n\to \infty} \left ( \frac{1}{\sqrt{n}\sqrt{n+1}} +\frac{1}{\sqrt{n}\sqrt{n+2}} + ... + \frac{1}{\sqrt{n}\sqrt{n+n}} \right ) \\

\lim_{n\to \infty} \sum_{i=1}^{n} \frac{1}{\sqrt{n}\sqrt{n+i}}\\

\lim_{n\to \infty} \sum_{i=1}^{n} \frac{1}{n\sqrt{1+i/n}} \\

\lim_{n\to \infty} \sum_{i=1}^{n} \frac{1}{\sqrt{1+i/n}} \frac{1}{n} ~~~~~~~~~~~~~~~~~~~~(1)$$.

Now, how can we write the last expression into an integral. From the actual expression of Riemann sum (the one I have written at top) we can infer by direct comparison that $$ \frac{(b-a)}{n} \rightarrow \frac{1}{n}$$ and hence $b-a = 1$ and from the

**input**of the function in Riemann sum expression we can infer, again, by direct comparison that $$\left( a +\frac{(b-a)}{n}i \right) \rightarrow \left(1 + i/n \right)$$ and therefore, I conclude that $a=1$ and therefore $b=2$. But what is the function? What is $f(x)$? Is $f(x) = \frac{1}{x}$, here I have assumed that in $\frac{1}{\sqrt{ 1+i/n}} $ the whole thing inside the radical corresponds to $x$.

Please help me. I think I have got some weakness in my foundations so please explain me everything in your answer.