$X_n = \sqrt[k]{n^{p}+an^{q}+1}-\sqrt[k]{n^{p}+bn^{q}+1} $

Question

For what given p and q below sequence is bounded?

$X_n = \sqrt[k]{n^{p}+an^{q}+1}-\sqrt[k]{n^{p}+bn^{q}+1} $ where $0\leq q<p$ and $a\ne b$

My try

$(1+n)^{p}=1+n^{p}+\sum_{i=1}^{p-1}\binom{p}{i}n^{i}$

$an^{q}=\sum_{i=1}^{p-1}\binom{p}{i}n^{i}$

$a=\frac{\sum_{i=1}^{\frac{p}{k}-1}\binom{\frac{p}{k}}{i}n^{i}}{n^{q}}$ we have same equality for $b$

we need that $|X_n|\leq M$

$M \geq (1+n)^{p/k}-(1+n)^{p/k} = an^{q} - bn^{q} = $

I don't know how to proceed after this to get something simple. maybe someone can put me into right direction?

Answer 1

If $q\gt0$, for every $c$, $$\sqrt[k]{n^{p}+cn^{q}+1}=n^{p/k}\,\sqrt[k]{1+cn^{q-p}+o(n^{q-p})}=n^{p/k}\,\left(1+ck^{-1}n^{q-p}+o(n^{q-p})\right),$$ hence $$x_n\sim n^{p/k}\,(a-b)k^{-1}n^{q-p}=\Theta(n^c),\qquad c=q-p+p/k,$$ and $(x_n)$ is bounded if and only if $c\leqslant0$, that is, $$kq\leqslant(k-1)p.$$