Converting a limit into an integral.

Question

I have to convert this limit $$ \lim_{n\to \infty} \left ( \frac{1}{\sqrt{n}\sqrt{n+1}} +\frac{1}{\sqrt{n}\sqrt{n+2}} + ... + \frac{1}{\sqrt{n}\sqrt{n+n}} \right )$$
The expression of limit of Riemann Sum that I know is $$ \lim_{k\to \infty} \sum_{I=1}^{k} f \left (a + \frac{(b-a)}{k}i\right ) \frac{b-a}{k}$$ we can write our limit as $$ \lim_{n\to \infty} \sum_{i=1}^{n} \frac{1}{\sqrt{n+i}} \frac{1}{\sqrt n}$$
By direct comparison we can say that $$ \frac{b-a}{k} \rightarrow \frac{1}{\sqrt n} \\ f\left (a+ \frac{(b-a)}{k}i \right) \rightarrow \frac{1}{\sqrt{n+i}}$$.

When $i=1$ we have $ f\left ( a + \frac{(b-a)}{k}\right) \rightarrow \frac{1}{\sqrt {n+1}}$. I can't seem to be getting $a$ and $b$. How can I proceed after this?

Please help me out here, also if there lies any mistake or if there are any other method then please write it.

EDIT:- When we write $$ \lim_{n\to \infty} \sum_{I=1}^{n} f \left(a+ \frac{(b-a)}{n} i \right) \frac{(b-a)}{n} = \int_{a}^{b} f(x) dx $$ I know that $\frac{b-a}{n}$ became $dx$ but what became the input $x$. Does the whole thing $a+ \frac{(b-a)}{n} i$ became $x$ or the $i$ which is varying became $x$.

The main source of my problem is that when @MarkViola wrote $$ f\left (a+ \frac{b-a}{n}i \right) \rightarrow \frac{1}{\sqrt{1+i/n}} \\ \implies f(x) = \frac{1}{\sqrt{1+x}}$$

I couldn’t understand that what is the input in $\frac{1}{\sqrt{1+i/}}$. I can see that the function is making everything into a square root and then reciprocating it but what is the input. Is the function replacing it’s input to $\frac{1}{\sqrt{1+i/n}}$ by the $i$ or the whole expression in the radical sign is the input

Answer 1

Note that we can write

$$\sum_{i=1}^n \frac{1}{\sqrt n\sqrt{n+i}}=\frac1n\sum_{i=1}^n\frac1{\sqrt{1+\left(\frac in\right)}}\tag 1$$

For the integral $\int_a^b f(x)\,dx$, the Riemann sum is written

$$ \frac{b-a}n\sum_{i=1}^n f\left(a+\frac{(b-a)i}{n}\right)\tag2$$

Comparing $(1)$ to $(2)$ we see that we can set $a=0$, $b=1$, and $f(x)=\frac1{\sqrt{1+x}}$.

Can you finish now?

NOTE:

It is important to note that this is not the unique answer. We could, for example, have $a=1$, $b=2$, and $f(x)=\frac1{\sqrt x}$.

Answer 2

The sum is $$\frac1n\sum_{k=1}^n\frac1{\sqrt{1+k/n}}$$ which looks like a Riemann sum for $1/\sqrt{1+x}$.

Answer 3

It's $$\lim_{n\rightarrow+\infty}\sum_{k=1}^n\left(\frac{1}{\sqrt{1+\frac{k}{n}}}\cdot\frac{1}{n}\right)=\int_0^1\frac{1}{\sqrt{1+x}}dx=2\sqrt2-2$$