For which $t\in\mathbb{R}$ does the sequence converge?

Question

I have to find some $t\in\mathbb{R}$ for which the following sequence

$$(b_n)_{n\in\mathbb{N}} \text{ with } b_n:=\sqrt{n^2+n^t}-n$$

does converge. I've found out that $b_n$ has the following limits:

$$ \lim_{n\to\infty} b_n = \begin{cases} 0,&\ t<1\\ \frac{1}{2},&\ t=1\\ \infty, &\ t\geq 2 \end{cases} $$

How can I compute the limit for the case $t\in(1,2)$?

Answer 1

Write $t=2-\delta$, $0<\delta<1$, then $\sqrt{n^2+n^t}-n=n\sqrt{1+n^{-\delta}}-n=n\big(\sqrt{1+n^{-\delta}}-1\big)=n\frac{\big(\sqrt{1+n^{-\delta}}-1\big)\big(\sqrt{1+n^{-\delta}}+1\big)}{\sqrt{1+n^{-\delta}}+1}=n\frac{n^{-\delta}}{\sqrt{1+n^{-\delta}}+1}=\frac{n^{1-\delta}}{\sqrt{1+n^{-\delta}}+1}$

The denominator $\to2$ as $n\to\infty$ whilst the numerator $\to\infty$ as $n\to\infty$, so the expression $\to\infty$.

Answer 2

We have $$b_n = \sqrt{n^2+n^t}-n = \frac{(\sqrt{n^2+n^t}-n)(\sqrt{n^2+n^t}+n)}{(\sqrt{n^2+n^t}+n)} = \frac{n^{t-1}}{(\sqrt{1 + n^{t-2}} + 1)} \to \frac{"\infty "}{2}$$