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.
Answer
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?
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