How I can calculate the limit of this sequence How I can calculate this limit:
$$\lim_{n→\infty}(\alpha^{1/(n \cdot \ln n)}-1)^{1/n}$$
where $2<α<3$ is a real number. Or at least how I can see if it is convergente sequence or not.
 A: Let $\varphi(n)=\left(f(n)-1\right)^{\frac{1}{n}}=\exp\left[\frac{\log\left(f(n)-1\right)}{n}\right]$ where $f(n)=\alpha^{\frac{1}{n\log n}}\to 1$ as $n\to\infty$.
We have 
$$
\lim_{n\to\infty}\frac{\log (f(n)-1)}{n}\stackrel{\tiny l'Hospital}{=}\lim_{n\to\infty}\frac{f'(n)}{f(n)-1}=\lim_{n\to\infty}\frac{-\frac{\log\alpha(1+\log n)}{n^2\log^2 n}f(n)}{f(n)-1}=\lim_{n\to\infty}\frac{-\log\alpha}{n^2\log n}=0
$$
so that
$$
\lim_{n\to\infty}\varphi(n)=1.
$$
EDIT: if you can use Tayolor's expansion, you have $f(n)\sim 1+\frac{\log\alpha}{n\log n}$ so that $\log\varphi(n)\sim\frac{\log\log\alpha-\log n-\log\log n}{n}\to 0$ and $\varphi(n)\to 1$.
A: Multiplying by $\displaystyle \left(\frac{1/(n\log n)}{1/(n\log n)}\right)^{1/n} $ we get $$\lim_{n\to \infty}\left(\frac{1}{n\log n}\right)^{1/n} \left(\frac{\alpha^{1/(n\log n)}-1}{1/(n\log n)}\right)^{1/n}=\\\lim_{n\to \infty}\left(\frac{1}{n\log n}\right)^{1/n}\cdot \lim_{n\to \infty}\left(\frac{\alpha^{1/(n\log n)}-1}{1/(n\log n)}\right)^{1/n}.$$
We analise the latter limits separately. 
For large enough $n$ we have $$\left(\frac{1}{n^2}\right)^{1/n}<\left(\frac{1}{n\log n}\right)^{1/n}<\left(\frac{1}{n}\right)^{1/n}, $$ so by the squeeze theorem, $$\lim_{n\to \infty}\left(\frac{1}{n\log n}\right)^{1/n}=1. $$
As to the second limit, let $\displaystyle m=\frac{1}{n\log n}$ to obtain $$\lim_{m\to 0}_{n\to \infty} \left(\frac{\alpha^m-1}{m}\right)^{1/n}=\lim_{n\to \infty}(\log\alpha)^{1/n}=1.$$
Thus we conclude $$\lim_{n→\infty}\left(\alpha^{1/(n \cdot \log n)}-1\right)^{1/n}=1. $$
A: It appears to be $1$, judging from the graphs of
$$y=(\alpha^{-x\ln x}-1)^x$$
with $x$ ranging from $0.001$ to $.1$ in steps of $0.001$. (Think of $x$ as $1/n$.) Tried with big and small values of $\alpha$ (ranging between $1.000000001$ and $10^{20}$).
The question now is, how to prove it...
