Without Taylor series $\lim\limits_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) $ Is it possible to compute it without Taylor series?
$$\lim_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) 
$$
Maybe you try your luck without computational software. 

We know the famous result:
$$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+o(1)$$
so we find
$$\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1=\exp(\frac{1}{n}\log\left(\log(n)+\gamma+o(1)\right))-1\sim_\infty\frac{\log\log n}{n}$$
hence
$$\lim_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) =\lim_{n\to\infty}\frac{n}{\ln \ln n}\frac{\log\log n}{n}=1.$$
