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. 
 A: 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
$$\lim_{n\to\infty}\frac{n}{\ln \ln n}\cdot \left(\sqrt[n]{1+\frac{1}{2}+\cdots+\frac{1}{n}}-1\right) =\lim_{n\to\infty}\frac{n}{\ln \ln n}\frac{\log\log n}{n}=1.$$
A: 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 $(3)$.
A: $ 1 + \frac  1 2 + ... + \frac 1 n = \log n + O(1)$. So 
$$ \left( 1 + \frac  1 2 + ... + \frac 1 n \right)^{\frac 1 n} = \exp \left( \frac 1 n \log(\log n + O(1)) \right) = 1 +  \frac 1 n \log(\log n) + o \left( \frac {\log(\log n)} n  \right). $$
So the answer to your problem is $1$ (but I used Taylor series...)
A: Doing this any way but Taylor series is in my mind, insane.  But still, you can use L'Hopital.  The trick is to express
$$[\log{n}]^{1/n} = \exp{\left ( \frac{\log{\log{n}}}{n}\right)}$$
So what is desired is
$$\lim_{n \rightarrow \infty} \frac{n \left [ \exp{\left ( \frac{\log{\log{n}}}{n}\right)} -1 \right ]}{\log{\log{n}}}$$
Using Taylor series, you can eyeball this and see that the limit is $1$.  L'Hopital is the next best thing, however, but it does get messy.  You have to take derivatives of numerator and denominator separately, as we have a $0/0$ situation.  After some messy algebra, we get the equivalent limit:
$$\lim_{n \rightarrow \infty} \left [ \left ( \exp{\left ( \frac{\log{\log{n}}}{n}\right)} -1 \right ) n \log{n} - \exp{\left ( \frac{\log{\log{n}}}{n}\right)} \log{n}\,\log{\log{n}} \right ] + \exp{\left ( \frac{\log{\log{n}}}{n}\right)}$$
Again, you can see that the term in the brackets goes to zero in this limit if you just examine the first term in the series.  But, if you are holding fast to ignorance of the behavior of $\exp{z}-1 \sim z$ near the origin, then you could rearrange the term in the brackets so that you get several complicated terms of the for $0/0$. At this point, however, I will depart you in the knowledge that the series is just way, way, way easier.
A: At Fisrt, following inequality holds:
$$\ln n \le H_n \le \ln n+1$$
Where $H_n$ is Harmonic number. So, we consider following limits:
$$\lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{\ln n} -1) \quad\text{and}\quad \lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{1+\ln n} -1) $$
We apply L'Hôpital's rule to calculate this limit. Before applying L'Hôpital's rule, we check this limit satisfies conditions of L'Hôpital's rule.
At First, we prove that
$$\lim_{x\to\infty} \frac{\ln\ln x}{x}=0 \quad\text{and}\quad \lim_{x\to\infty} \sqrt[x]{\ln x} =1.$$
To prove these formulas is easy, and 
$$\lim_{x\to\infty} \frac{(\sqrt[x]{\ln x} -1)'}{(\ln\ln x/x)'}= \lim_{x\to\infty} \frac{\sqrt[x]{\ln x}\cdot (\ln\ln x/x)' }{(\ln\ln x/x)'}=1$$
So By L'Hôpital's rule we get
$$\lim_{x\to\infty} \frac{\sqrt[x]{\ln x} -1}{\ln\ln x/x}=\lim_{x\to\infty} \frac{(\sqrt[x]{\ln x} -1)'}{(\ln\ln x/x)'}= 1$$
And you can show that $$\lim_{n\to\infty} \frac{n}{\ln\ln n} (\sqrt[n]{1+\ln n} -1)=1.$$
A: Without Taylor series.
First we show that, for $0 \le x \le 1$, we have that
$$x \le e^x -1 \le x + ex^2$$
This can easily be shown by taking derivatives.
For instance, for the right side, let $f(x) = e^x - 1 - x - ex^2$.
We have that $f(0) = 0$.
Now $f'(x) = e^x - 1 -2ex$. Taking derivatives again, we can show that
$f'(x) \le 0$ for $0 \le x \le 1$ (basically $f'(0) = 0$ and $f''(x) \le 0$ for $0 \le x \le 1$)
Thus $f(x) \le 0$ for $0 \le x \le 1$.
Now just use any elementary proof that $\log (n-1) \le H_n \le \log n+ 1$ (say using integrals).
Let
$$ S_n = H^{1/n} -1 = e^{\frac{\log H_n}{n}} -1$$
Now using the above inequality (for sufficiently large $n$), we get that
$$ \frac{\log H_n}{n} \le S_n \le \frac{\log H_n}{n} + \frac{e\log^2 H_n}{n^2}$$
Now by mean value theorem, for any $c > 0$ we have that $\log \log (n+c) - \log \log n \le \frac{c}{n \log n}$ and thus $\lim_{n \to \infty} \frac{\log \log (n+c)}{\log \log n} = 1$.
Now it is easy to see that your sequence has the limit $1$.
A: A limit I use a bit, which is the inverse of $\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$:
$$
\lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)\tag{1}
$$
Therefore, for any constant $\alpha$, substituting $n\mapsto\frac{n}{\log(\alpha+\log(n))}$, we get
$$
\lim_{n\to\infty}\frac{n}{\log(\alpha+\log(n))}\left(x^{\log(\alpha+\log(n))/n}-1\right)=\log(x)\tag{2}
$$
Plugging $x=e$ into $(2)$ yields
$$
\lim_{n\to\infty}\frac{n}{\log(\alpha+\log(n))}\left((\alpha+\log(n))^{1/n}-1\right)=1\tag{3}
$$
Since $1+\frac12+\frac13+\dots+\frac1n\to\gamma+\log(n)$ and $\lim\limits_{n\to\infty}\frac{\log(\alpha+\log(n))}{\log(\log(n))}=1$, $(3)$ gives
$$
\lim_{n\to\infty}\frac{n}{\log(\log(n))}\left(\left(1+\frac12+\frac13+\dots+\frac1n\right)^{1/n}-1\right)=1\tag{5}
$$
A: We begin by noticing that the expression under root sign is $n$th Harmonic number, so rewrite the problem $$\lim_{n \to \infty} \frac {n} {\log \log n} \left(\sqrt [n] {H_n} - 1 \right).$$ Observe that $\log n < H_n < \log n + 1$.
Let $x = \log n$, and $x \to \infty$. This means that $$H_n^{1/n} = \exp \left(\frac {\log x} {\exp x}\right) + o \left (\frac {1} {\exp x}\right).$$ The problem becomes $$\lim_{x \to \infty} \exp \left(\frac {\log x} {\exp x}\right) + o \left(\frac {1} {\log x} \right) = \exp 0 + 0 = 1.$$
