Evaluate $\lim_{n\to \infty}{1+{1\over 2}+\cdots +{1\over n}\over (\pi ^n +e^n)^{1\over n}\log_e n}$ Evaluate $$\lim_{n\to \infty}{1+{1\over 2}+\cdots +{1\over n}\over (\pi ^n +e^n)^{1\over n}\log_e n}.$$
My attempt: $${1\over \pi}\lim_{n\to \infty}{1+{1\over 2}+\cdots +{1\over n}\over (1 +({e\over \pi})^n)^{1\over n}\log_e n}={1\over \pi}\lim_{n\to \infty}{1+{1\over 2}+\cdots +{1\over n}\over \log_e n}={1\over \pi}\times 1.$$
As $\lim_{n\to \infty}(1 +({e\over \pi})^n)^{1\over n}=1$.
Is there any mistake in this. If so please rectify this. Also any other way to solve this will be appreciated. Thanks in advance.
 A: Let $$L=\lim_{n\to \infty}{1+{1\over 2}+\cdots +{1\over n}\over (\pi ^n +e^n)^{1\over n}\log_e n}.$$
$$L=\lim_{n\to \infty} \frac{1+1/2+1/3+...+1/n}{\ln n}\lim_{n \to \infty} (\pi^n+e^n)^{-1/n}$$
By Euler's asynptotic result $1/1+1/2+1/3+1/4+..+1/n\sim \ln n$, we can write
$$L=\lim_{n\to \infty}(\pi^n+e^n)^{-1/n}.$$
Next by sqeeze law $$\pi^n<(\pi^n+e^n)<2\pi^n \implies 2^{-1/n}\pi^{-1} <(\pi^n+e^n)^{-1/n} < \pi^{-1}.$$
Hence $L=\frac{1}{\pi}.$
A: For
$$ a_{n} = \frac{H_{n}}{(\pi ^n +e^n)^{1\over n} \, \ln n}, $$
where $H_{n}$ is the harmonic number, it can be seen that
$$ H_{n} \approx \ln n + \gamma + \frac{1}{2 \, n} + \mathcal{O}\left(\frac{1}{n^2}\right) $$
and $\frac{e}{\pi} < 1$ which gives
$$ \left(\pi^n + e^n \right)^{1/n} = \pi \, \left(1 + \left(\frac{e}{\pi}\right)^n \right)^{1/n} = \pi \, \left( 1 + \frac{1}{n} \, \left(\frac{e}{\pi}\right)^n + \mathcal{O}\left(\frac{1}{n^2}\right) \right). $$
Now,
$$ a_{n} \approx \frac{1 + \frac{\gamma}{\ln n} + \frac{1}{2 \, n \, \ln n} + \mathcal{O}\left(\frac{1}{n^2}\right)}{\pi \, \left( 1 + \frac{1}{n} \, \left(\frac{e}{\pi}\right)^n + \mathcal{O}\left(\frac{1}{n^2}\right) \right) }. $$
Taking the desired limit leads to
$$ \lim_{n \to \infty} \, a_{n} = \frac{1}{\pi}. $$
