<s>[INST] Infinite series of $\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$ Find Value of  $$S=\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$$
$$S=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+4n+4}}$$
But here i cannot express above sum in form of $$\lim_{n \to \infty} \frac{1}{n} \times \sum_{r=1}^{n}f(\frac{r}{n})$$
 [/INST] On the interval $r\in[n^2,(n+2)^2]$ the terms $\frac{1}{\sqrt{r}}$ have a quite small variation: they are approximately all equal to $\frac{1}{n}$; since there are $4n+5$ terms, the limit is expected to be $\color{red}{4}$. 
In a more rigorous way,
$$ \frac{4n+5}{n+2}\leq \sum_{r=n^2}^{(n+2)^2}\frac{1}{\sqrt{r}} \leq \frac{4n+5}{n} $$
and the previous claim follows by squeezing.
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