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. 

Here is an idiosyncratic proof:
First, applying the squeezing lemma to the inequality $ 1 \leq (H_{n})^{1/n} \leq n^{1/n} $, we know that
$$\lim_{n\to\infty} (H_{n})^{1/n} = 1. \tag{1} $$
Next, as tetori pointed out, we have
$$ \log n \leq H_n \leq 1 + \log n. \tag{2} $$
From this inequality, we have
$$ \log \log n \leq \log H_n \leq \log(1 + \log n) = \log\log n + \log \left( 1 + \frac{1}{\log n} \right) $$
In particular, dividing each sides by $\log \log n$ and taking $n \to \infty$ gives
$$\lim_{n\to\infty} \frac{\log H_n}{\log \log n} = 1. \tag{3} $$
Now it is plain to observe that
\begin{align*}
\frac{n}{\log\log n} \{ (H_n)^{1/n} - 1 \}
&= \frac{1}{\log\log n} \int_{1}^{H_n}  \frac{x^{1/n}}{x} \, dx.
\end{align*}
Since
\begin{align*}
\log H_n 
&= \int_{1}^{H_n} \frac{1}{x} \, dx
 \leq \int_{1}^{H_n}  \frac{x^{1/n}}{x} \, dx 
 \leq \int_{1}^{H_n}  \frac{(H_n)^{1/n}}{x} \, dx
 = (H_n)^{1/n} \log H_n,
\end{align*}
we have
$$ \frac{\log H_n}{\log\log n}
\leq \frac{n}{\log\log n} \{ (H_n)^{1/n} - 1 \}
\leq (H_n)^{1/n}  \frac{\log H_n}{\log\log n} .$$
Therefore, taking $n\to\infty$ we obtain the desired limit by both $(1)$ and $(3)$.
