$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$

Question

$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$

I can change the form of this limit saying that $n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2(1+\frac{2}{n}+\frac{1}{n^2})})=n^{\alpha+\frac{2}{5}} (\sqrt[5]{1+\frac{1}{n}}-\sqrt[5]{1+\frac{2}{n}+\frac{1}{n^2}})$ Then $\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases}+ \infty & \alpha>-\frac{2}{5}\\ 2 & \alpha=-\frac{2}{5}\\0 & \alpha<-\frac{2}{5}\end{cases}$ Is it right?

Answer 1

By binomial expansion

• $\sqrt[5]{n^2+n}=\sqrt[5]{n^2}\sqrt[5]{1+1/n}=\sqrt[5]{n^2}(1+\frac1{5n}+o(1/n))$
• $\sqrt[5]{n^2+2n+1}=\sqrt[5]{n^2}\sqrt[5]{1+2/n+1/n^2}=\sqrt[5]{n^2}(1+\frac2{5n}+o(1/n))$

then

$$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=-\frac{n^\alpha\sqrt[5]{n^2}}{5n}+o\left(\frac{n^\alpha\sqrt[5]{n^2}}{n}\right)=-\frac15 n^{\alpha-\frac35}+o\left(n^{\alpha-\frac35}\right)$$

therefore

$$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})= \begin{cases} - \infty & \alpha>\frac{3}{5}\\ -\frac15 & \alpha=\frac{3}{5}\\ 0 & \alpha<\frac{3}{5}\end{cases}$$

Answer 2

It's simpler if you consider the limit you get by (formally) substituting $n=1/x$: $$ \lim_{x\to0^+}\frac{1}{x^{\alpha}}\frac{\sqrt[5]{1+x}-\sqrt[5]{1+2x+x^2}}{\sqrt[5]{x^2}} $$ If this limit exists (finite or infinite), then it's the same as the limit of your sequence. Now use Taylor: $$ \frac{1+\dfrac{1}{5}x-1-\dfrac{2}{5}x+o(x)}{x^{\alpha+2/5}}= \frac{-\dfrac{1}{5}+o(1)}{x^{\alpha-3/5}} $$ Now it should be easy.