Find the given limit
$$\lim_{n\to\infty}\frac{1+1/2+1/3+\ldots+1/n}{(\pi^{n}+e^{n})^{1/n}\ln n}$$
I'm able to find one part in denominator of this limit i.e. $\lim_{n\to \infty} (\pi ^{n} + e^{n})^{1/n} = \pi$ So there will be a $\pi$ in the denominator of the answer. How to find the rest part$?$
Note that $1+1/2+1/3+...+1/n \sim \gamma+\ln n$, where $\gamma=0.577\ldots$, is the Euler–Mascheroni constant. So your limit becomes $$ L =\lim_{n\rightarrow \infty} \frac{\ln n+\gamma}{\pi \ln n} = \lim_{n \rightarrow \infty} \left( \frac{1}{\pi}+ \frac{\gamma}{\pi\ln n} \right) =\frac{1}{\pi}. $$
Let $a_n := \sum_{k = 1}^{n} \frac{1}{k}$. Since $b_n := \sqrt[n]{\pi^n + e^n} \ln(n)$ is a strictly monotone sequence diverging to $\infty$, by the theorem of Stolz-Cesà ro we have \begin{align} \lim_{n \to \infty} \frac{a_n}{b_n} & = \lim_{n \to \infty} \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} \\ & = \lim_{n \to \infty} \frac{1}{(n + 1) \big[ \sqrt[n + 1]{\pi^{n + 1} + e^{n + 1}} \ln(n+1) - \sqrt[n]{\pi^{n} + e^{n}} \ln(n)\big]} = \frac{1}{\pi} \end{align} To see this we show that $$\lim_{n \to \infty} \frac{1}{(n + 1) \left[\left(1 + \left(\frac{e}{\pi}\right)^{n + 1}\right)^{\frac{1}{n + 1}} \ln(n+1) - \left( 1 + \left(\frac{e}{\pi}\right)^{n}\right)^{\frac{1}{n}} \ln(n)\right]} = 1. $$ This follows from the $\left( 1 + \left(\frac{e}{\pi}\right)^{n}\right)^{\frac{1}{n}} \xrightarrow{n \to \infty} 1$ (thanks to op for pointing that out) and $$(n + 1)( \ln(n + 1) - \ln(n) ) \xrightarrow{n \to \infty} 1.$$
