1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.