I have to find the infinitesimal order of a function $f(n)$ as $n \to \infty$, this is what I did:
$$ \begin{split} f(n) &= \sqrt{n+1} - \sqrt{n} + \frac{1}n \\ &= \frac{n\sqrt{n+1} - n\sqrt{n} + 1}{n} \\ &= \frac{\sqrt{n^3(1+\frac{1}{n^2})}-\sqrt{n^3}+1}{n} \\ &= \frac{\sqrt{n^3}-\sqrt{n^3}+1}{n} \\ &= \frac{1}{n} \end{split} $$ So the infinitesimal order is: 1
Is it correct? Or should I have used some notable limit, taylor or asymptomatic approximation.
On
I would consider the following instead: $$ \begin{split} \sqrt{n+1} - \sqrt{n} &= \left(\sqrt{n+1} - \sqrt{n}\right) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\ &= \frac{n+1-n}{\sqrt{n+1} + \sqrt{n}} \\ &= \frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &= \Theta\left(n^{-1/2}\right) \end{split} $$ therefore, $$ \begin{split} f(n) &= \sqrt{n+1} - \sqrt{n} + \frac1n \\ &= \frac{1}{\sqrt{n+1} + \sqrt{n}} + \frac1n \\ &= \Theta\left(n^{-1/2} + n^{-1}\right) \\ &= \Theta\left(n^{-1/2}\right) \end{split} $$
Generally it's easier to note that $$ \sqrt{n+1}=\sqrt{n}\sqrt{1+1/n}\sim \sqrt{n}\left(1+\frac{1}{2n}\right)=\sqrt{n}+\frac{1}{2\sqrt{n}}, $$ using the Taylor series for $\sqrt{1+x}$. Then $$ \sqrt{n+1}-\sqrt{n}+\frac{1}{n}\sim\frac{1}{2\sqrt{n}}+\frac{1}{n}, $$ with omitted terms $O(n^{-3/2})$. The error in your original calculation is that $n\sqrt{n+1}=\sqrt{n^3+n^2}=\sqrt{n^3(1+1/n)}$, not $\sqrt{n^3(1+1/n^2)}$.