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.
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.