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Calculate the limit $$ \lim_{n \to \infty} \frac{1+\frac{1}{2}+\ldots +\frac{1}{n}}{\left(\pi^n+e^n \right)^{\frac{1}{n}} \ln{n}} $$

I tried solving the question with L hospital rule but didn't get to the answer instead the question become more complex. Please solve the question or at least provide some hints. Thanks in advance

Matti P.
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newbie
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    Let $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$ (this is the $n$-th harmonic number). Do you know some bounds for $H_n$ involving the natural logarithm function (if not, try and come come up with some by considering integrals of the decreasing function $1/x$ over intervals like $[1,n]$. Or see https://math.stackexchange.com/questions/2832848/bounds-for-the-harmonic-k-th-partial-sum)? Also, try and show that $\lim\limits _{n\to\infty}\left(a^n + A^n\right)^{1/n} = A$, for any real numbers $0\le a \le A$. – Minus One-Twelfth Mar 28 '19 at 10:25

2 Answers2

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Hint: show that $\frac {1+\frac 1 2+...+\frac 1 n} {\log_e\,n }\to 1$ and $(\pi^{n}+e^{n})^{1/n})^{-1} \to \frac 1 {\pi}$ so the answer is $\frac 1 {\pi}$.

Kavi Rama Murthy
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You have, when $n$ tends to $+\infty$, $$\frac{\sum_{k=1}^n \frac{1}{k}}{(\pi^n + e^n)^{1/n} \ln(n)} \sim \frac{\ln(n)}{(\pi^n + e^n)^{1/n} \ln(n)} = \frac{1}{(\pi^n + e^n)^{1/n} } = \frac{1}{\pi \left(1 + \left(\frac{e}{\pi}\right)^n\right)^{1/n} }$$

and $$\left(1 + \left(\frac{e}{\pi}\right)^n\right)^{1/n} = \exp \left(\frac{1}{n} \ln \left(1 + \left(\frac{e}{\pi}\right)^n \right) \right) \rightarrow 1$$

So the limit is $$\frac{1}{\pi}$$

TheSilverDoe
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