ca.analysis and odes - Growth of the "cube of square root" function

Here is a proof that $|g(n)|le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.

If $f(n)=n^{3/2}$ without the floor, then $g(n)sim frac{3}{4sqrt n}$, so it is positive and tends to 0. When you replace $n^{3/2}$ by its floor, $g(n)$ changes by at most 2, hence the only chance for failure is to have $g(n)=2$ when the fractional parts of $n^{3/2}$ and $(n+2)^{3/2}$ are very small and the fractional part of $(n+1)^{3/2}$ is very close to 1 (the difference is less than $frac{const}{sqrt{n}}$).

Let $a,b,c$ denote the nearest integers to $n^{3/2}$, $(n+1)^{3/2}$ and $(n+2)^{3/2}$. Then $c-2b+a=0$ because it is an integer very close to $(n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}$. Denote $m=c-b=b-a$. Then $(n+1)^{3/2}-n^{3/2}<m$ and $(n+2)^{3/2}-(n+1)^{3/2}>m$. Observe that
$$frac{3}{2}sqrt{n}<(n+1)^{3/2} - n^{3/2} < frac{3}{2}sqrt{n+1}$$
(the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore
$$frac{3}{2}sqrt{n} < m < frac{3}{2}sqrt{n+2}$$
or, equivalently,
$$n < frac49 m^2 < n+2.$$
If $m$ is a multiple of 3, this inequality implies that $n+1=frac49 m^2=(frac23m)^2$, then $(n+1)^{3/2}=(frac23m)^3$ is an integer and not slightly smaller than an integer as it should be. If $m$ is not divisible by 3, then
$$n+1 = frac49 m^2 + r$$
where $r$ is a fraction with denominator 9 and $|r|<1$. From Taylor expansion
$$f(x+r) = f(x) +r f'(x) +frac12 r^2 f''(x+r_1), 0<r_1<r,$$
for $f(x)=x^{3/2}$, we have
$$(n+1)^{3/2} = (frac49 m^2 + r)^{3/2} = frac8{27}m^3 + mr + delta$$
where $0<delta<frac1{4m}$.
This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.

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