Evaluating $\lim_{n \to \infty}\;\frac{ 1+ \frac{1}{2}+\cdots+\frac{1}{n}}{ (\pi^n + e^n)^{1/n} \ln{n} }$ I have
$$\lim_{n \to \infty}\;\frac{ 1+ \frac{1}{2}+\cdots+\frac{1}{n}}{ (\pi^n + e^n)^{1/n} \ln{n} }$$
How should I proceed? Is there a way to use integration as limit of a sum here?
 A: First of all:
$$\lim_{n\to \infty} (\pi^n+e^n)^{\frac{1}{n}} = \lim_{n\to \infty} \pi \left[1+\left(\frac{e}{\pi}\right)^n\right]^{\frac{1}{n}} = \pi$$
since $\pi > e$. So the limit equals:
$$\frac{1}{\pi} \lim_{n\to \infty} \frac{1}{\ln n} \left(1+\frac{1}{2}+\ldots +\frac{1}{n}\right)$$
Intuitively the inner limit should be one because $H_n \sim \ln n + \gamma$, but we can also prove it with Cesaro-Stolz:
$$\lim_{n\to \infty} \frac{1}{\ln n} \left(1+\frac{1}{2}+\ldots +\frac{1}{n}\right) = \lim_{n\to \infty}\frac{1}{n+1} \ln\frac{n}{n+1} = \lim_{n\to \infty}\frac{1}{\ln \left(1+\frac{1}{n}\right)^{n+1}} = \frac{1}{\ln e} = 1$$
The final result is $\dfrac{1}{\pi}$.
A: Hint:
$1+ \frac{1}{2}+\cdots+\frac{1}{n}$ is an upper Riemann sum for the integral $\;\int_1^{n+1}\frac {\mathrm d x}x=\ln(n+1)\;$ and $\;\frac{1}{2}+\cdots+\frac{1}{n}\;$  is a lower Riemann sum for the integral $\;\int_1^n \frac {\mathrm d x}x=\ln n$, so
$$\frac{\ln(n+1)}{ (\pi^n + e^n)^{1/n} \ln{n}}\le \frac{ 1+\frac{1}{2}+\cdots+\frac{1}{n}}{(\pi^n + e^n)^{1/n} \ln{n}}\le\frac{1+\ln n}{ (\pi^n + e^n)^{1/n} \ln{n}}.$$
Now $\frac{\ln(n+1)}{\ln{n}}$ and $\frac{1+\ln n}{\ln{n}}$ both tend to $1$ as $n\to\infty$, so there remains to find
$$\lim_{n\to\infty}\frac{1}{(\pi^n + e^n)^{1/n}}.$$
