Limit of geometric series becoming logarithm?

Question

Got this equation from WolframAlpha (Limit[Sum[4^(p/n)/n,{p,1,n}],n->∞]): $$\lim_{n \to \infty}\sum_{p=1}^n\frac{4^\frac{p}{n}}{n}=\lim_{n \to \infty}\frac{3*\sqrt[n]{4}}{(\sqrt[n]{4}-1)n}=\frac{3}{2\ln(2)}$$

Please explain the steps from the leftmost expression to the rightmost expression.

For context: I am looking for an alternative way to calculate change in function's value from one point to another (without using circular definite integral), and this came from Riemann sum of the function's derivative. Even though this is exactly the answer I need, I have no clue how n-th root turned into logarithm.
I went through university Calculus course, which used Openstax textbooks. As I understand, these textbooks have only a handful of chapters on series, outlining basics. What books could be useful for learning more of Calculus?

Answer 1

If $n,p\in\Bbb N$, then\begin{align}\sum_{p=1}^n\frac{4^{p/n}}n&=\frac1n\sum_{p=1}^n\sqrt[n]4^p\\&=\frac1n\cdot\frac{\sqrt[n]4\left(\sqrt[n]4^n-1\right)}{\sqrt[n]4-1}\\&=\frac{3\sqrt[n]4}{n\left(\sqrt[n]4-1\right)},\end{align}and therefore$$\lim_{n\to\infty}\sum_{p=1}^n\frac{4^{p/n}}n=\lim_{n\to\infty}\frac{3\sqrt[n]4}{n\left(\sqrt[n]4-1\right)}=\frac3{\log4},$$since, for each $x>0$, $\lim_{n\to\infty}\sqrt[n]x=1$ and $\lim_{n\to\infty}n\left(\sqrt[n]x-1\right)=\log x$. More generally,$$x\in(0,\infty)\setminus\{1\}\implies\lim_{n\to\infty}\sum_{p=1}^n\frac{x^{p/n}}n=\frac{x-1}{\log x}.$$