Find the supremum and infimum of the set $S = \{ \sqrt {n^2 + 1} - n: n \in \mathbb{N}\}$

Question

Find the supremum and infimum of the set $S = \{ \sqrt {n^2 + 1} - n: n \in \mathbb{N} \}.$ I know that the supremum is $\sqrt{2} - 1$ but what about the infimum is it $0$?

Could anyone tell me if I am right or wrong?

Answer 1

The perennial classic $$ \sqrt{n^2+1}-n = \frac{1}{\sqrt{n^2+1}+n} $$ should convince you that you're correct: it's clear that this is a decreasing function of $n$, and smaller than $1/(2n)$, which also tends to zero.

Answer 2

More generally, if $f(n) \to \infty$ and $\dfrac{g(n)}{f(n)} \to 0$, then

$\begin{array}\\ \sqrt{f^2(n)+g(n)}-f(n) &=(\sqrt{f^2(n)+g(n)}-f(n))\dfrac{\sqrt{f^2(n)+g(n)}+f(n)}{\sqrt{f^2(n)+g(n)}+f(n)}\\ &=\dfrac{(f^2(n)+g(n))-f^2(n)}{\sqrt{f^2(n)+g(n)}+f(n)}\\ &=\dfrac{g(n)}{\sqrt{f^2(n)+g(n)}+f(n)}\\ &\to 0\\ \end{array} $

Your case is $f(n) = n$ and $g(n) = 1$.

Note that if $\lim_{n \to \infty} \dfrac{g(n)}{f(n)}= L $, then $\sqrt{f^2(n)+g(n)}-f(n) \to \frac12 L $.

For example, if $f(n) = g(n) = n$, then $\sqrt{n^2+n}-n \to \frac12 $.