Doubt on a limit with parameter

Question

I have to study the value of the following limit: $$\lim_{n \to +\infty} n^{\alpha}(\sqrt[n]{n+1} - 1)$$ for $\alpha \in \mathbb{R}$. So I made the substitution $\displaystyle x = \frac{1}{n}$ and got the limit: $$\lim_{x \to 0^+} \frac{1}{x^{\alpha}}\left(\left(\frac{1}{x} +1\right)^{\displaystyle\frac{1}{x}}-1\right) = \lim_{x \to 0^+} \frac{1}{x^{\alpha}}\left(e^{\displaystyle\frac{1}{x}\log{\left(\frac{1}{x}+1\right)} }-1\right)$$ now I would like to expand in Taylor series, but I can't because the internal function doesn't vanish, how could I go on?

Answer 1

We recall that $\dfrac{\ln(n+1)}{n}\rightarrow 0$ $(n\rightarrow\infty)$. Hence $\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{\ln(n+1)}{n}}{\sqrt[n]{n+1}-1}=$

$$\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{\ln(n+1)}{n}}{e^{\dfrac{\ln(n+1)}{n}}-1}=1$$ So it suffices to determine $$\lim\limits_{n\rightarrow\infty}\dfrac{\ln(n+1)}{n^{1-\alpha}}$$ We then recall that L'Hopital rule implies that for fixed $t$, $$\lim\limits_{x\rightarrow+\infty}\dfrac{\ln(x+1)}{x^{t}}=\begin{cases} 0 & t>0 \\ +\infty& t\leq 0 \end{cases}$$

Answer 2

If you want to make a series expansion, consider $$A=\sqrt[n]{n+1}\quad \implies \quad \log(A)=\frac 1 n \log(n+1)=\frac 1 n\left(\log \left(1+\frac{1}{n}\right)+\log (n)\right)$$ $$\log(A)=\frac{\log (n)}{n}+\frac{1}{n^2}-\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$ $$A=e^{\log(A)}=1+\frac{\log (n)}{n}+\frac{\log ^2(n)+2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$ n^{\alpha}(\sqrt[n]{n+1} - 1)=n^{\alpha-1}\log(n)+\frac 12n^{\alpha-2}(\log^2(n)+2)+\cdots$$