Evaluate $\lim\limits_{n\to\infty} \frac{3+\sqrt{3}+\sqrt[3]{3}+\dots+\sqrt[n]{3}-n}{\ln n}$ 
Evaluate $$\lim\limits_{n\to\infty} \frac{3+\sqrt{3}+\sqrt[3]{3}+\dots+\sqrt[n]{3}-n}{\ln n}.$$

We haven't been taught how to do asymptotes, I think this problem should be solved using Cesaro-Stolz; $$\lim\limits_{n\to\infty} \frac{a_n}{b_n}=\lim\limits_{n\to\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
Note that this is my second question on this site.It's the first year i learned about limits so don't expect that this problem solves in a hard way.
 A: Yes, you are correct, we can use Stolz-Cesaro theorem:
$$\lim_{n\to\infty} \frac{\sum_{k=1}^n(\sqrt[k]{3}-1)}{\ln(n)}\stackrel{SC}{=}\lim_{n\to\infty} \frac{\sqrt[n+1]{3}-1}{\ln (n+1)-\ln(n)}=
\lim_{n\to\infty} \frac{e^{\left(\frac{\ln(3)}{n+1}\right)}-1}{\ln\left(1+\frac{1}{n}\right)}.$$
Now note that (see Arnaud Mortier's comment below)
$$\lim_{t\to 0}\frac{e^{t}-1}{t}=\lim_{t\to 0}\frac{e^{t}-e^0}{t-0}=(e^t)'_{t=0}=e^0=1$$
and
$$
\lim_{t\to 0}\frac{\ln(1+t)}{t}=1=\lim_{t\to 0}\frac{\ln(1+t)-\ln(1+0)}{t-0}=(\ln(1+t))'_{t=0}=\frac{1}{1+0}=1.$$
Can you take it from here?
A: 
I thought it might be of interest to present an approach that does not rely on the Stolz-Cesaro Theorem, but rather uses only (i) a pair of inequalities of the exponential function, (ii) a result from the integral test, and (iii) the squeeze theorem.  To that end, we present two primers with references to their proofs.


PRIMER $(1)$:
In THIS ANSWER, I showed using on the limit definition of the integral test and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le\frac{1}{1-x}} \tag1$$
for $x<1$.

PRIMER $(2)$:
As a result from the integral test (SEE THIS), we have for a function $f(x)$ that is positive valued and monotonically decreasing for $x\ge N$
$$\bbox[5px,border:2px solid #C0A000]{\int_N^{n+1}f(x)\,dx\le \sum_{k=N}^n f(k)\le f(N)+\int_N^n f(x)\,dx}\tag2$$

PROBLEM SOLUTION:
Equipped with Primers $(1)$ and $(2)$, we are ready to proceed.  
We begin with the function $f(x)=\sqrt[x]{3}-1=e^{\frac1x\log(3)}-1$ and write the sum of interest as $\sum_{k=1}^n f(k)=f(1)+\sum_{k=2}^n f(k)=2+\sum_{k=2}^n f(k)$.  
Inasmuch as $\lim_{n\to \infty}\frac{2}{\log(n)}=0$, we analyze the limit $\lim_{n\to \infty}\frac{\sum_{k=2}^nf(k)}{\log(n)}$.
Applying the left-hand sides of $(1)$ and $(2)$, we see that 
$$\begin{align}
\sum_{k=2}^nf(k)&\ge \int_2^{n+1}f(x)\,dx\\\\
&=\int_2^{n+1}(e^{\frac1x \log(3)}-1)\,dx\\\\
&\ge \int_2^{n+1}\frac{\log(3)}{x}\,dx\\\\
&=\log(3)(\log(n+1)-\log(2))\tag3
\end{align}$$
Applying the right-hand sides of $(1)$ and $(2)$, we see that 
$$\begin{align}
\sum_{k=2}^nf(k)&\le (\sqrt3-1)+\int_2^{n}f(x)\,dx\\\\
&=(\sqrt 3-1)+\int_2^{n}(e^{\frac1x \log(3)}-1)\,dx\\\\
&\le (\sqrt 3-1)+\int_2^{n}\frac{\log(3)}{x-\log(3)}\,dx\\\\
&=(\sqrt 3-1)+\log(3)(\log(n-\log(3))-\log(2-\log(3)))\tag4
\end{align}$$
Dividing the expressions in $(3)$ and $(4)$ by $\log(n)$, and applying the squeeze theorem, we obtain the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\frac{\sum_{k=1}^n(\sqrt[k]{3}-1)}{\log(n)}=\log(3)}$$
And we are done!
A: As noted already you are correct and by Stolz-Cesaro theorem:
$$\lim_{n\to\infty} \frac{\sum_{k=1}^n(\sqrt[k]{3}-1)}{\ln(n)}\stackrel{SC}{=}\lim_{n\to\infty} \frac{\sqrt[n+1]{3}-1}{\ln\left(1+\frac{1}{n}\right)}=\ln 3$$
indeed
$$\frac{\sqrt[n+1]{3}-1}{\ln\left(1+\frac{1}{n}\right)}= \frac{n\left(\sqrt[n+1]{3}-1\right)}{\ln\left(1+\frac{1}{n}\right)^n} \to \ln 3$$
since by definition of $e$
$$\ln\left(1+\frac{1}{n}\right)^n\to \ln e=1$$
and by Taylor's espansion
$$\sqrt[n+1]{3}=e^{\frac{\ln 3}{n+1}}=1+\frac{\ln 3}{n+1}+o\left(\frac1n\right)$$
$$n\left(\sqrt[n+1]{3}-1\right)=\frac{n}{n+1}\ln 3+o(1)\to\ln 3$$
