$$\lim_{n \to \infty} \frac{1}{n} \cdot\ln(3^\frac{n}{1}+3^\frac{n}{2}+\cdots+3^\frac{n}{n})$$ I tried to apply the squeeze theorem, but I can't manage to solve it.

  • $\begingroup$ Try moving $\frac1n$ inside the $\ln$ and then use AM-GM $\endgroup$ – Don Thousand Oct 31 '18 at 15:48
  • $\begingroup$ Hint: upper bounds for $n/1, n/2, n/3, \ldots, n/n$ are $n, n-1, n-2, \ldots, 1$. $\endgroup$ – Mees de Vries Oct 31 '18 at 15:51

The squeeze theorem is a good idea. $$ \ln(3^n)\leq \ln(3^\frac{n}{1}+3^\frac{n}{2}+\cdots+3^\frac{n}{n}) \leq \ln(n\cdot3^n) $$

  • $\begingroup$ Thank you! The result is $\ln 3$ $\endgroup$ – user69503 Oct 31 '18 at 16:14

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