Calculate $\lim\limits_{n \to \infty} \frac1n\cdot\log\left(3^\frac{n}{1} + 3^\frac{n}{2} + \dots + 3^\frac{n}{n}\right)$ 
Calculate $L = \lim\limits_{n \to \infty} \frac1n\cdot\log\left(3^\frac{n}{1} + 3^\frac{n}{2} + \dots + 3^\frac{n}{n}\right)$

I tried putting $\frac1n$ as a power of the logarithm and taking it out of the limit, so I got 
$$ L = \log\lim\limits_{n \to \infty} \left(3^\frac{n}{1} + 3^\frac{n}{2} + \dots + 3^\frac{n}{n}\right)^\frac1n $$
At this point I thought of the fact that $\lim\limits_{n \to \infty} \sqrt[n]{a_1^n+a_2^n+\dots+a_k^n} = max\{a_1, a_2, \dots,a_k\}$ but this won't be of any use here, I guess. How can I calculate this limit, please?
 A: You can use the Squeeze theorem. For example in this case you have
$$\log 3 =\frac{\log(3^n)}{n} \leq \frac{\log(3^n+3^{n/2}+\cdots+3^{n/n})}{n} \leq \frac{\log(n \cdot 3^n)}{n} = \frac{\log{n}}{n}+\log 3$$
Now it is not hard to see that both surrounding sequences converge to $\log 3$, so the limit of the middle is $\log 3$ as well.
A: Hint. $$0<3^{\frac n 2}+\cdots+3^{\frac n n}\le(n-1)3^{\frac n 2}=\frac{3^n}{3^{\frac n 2}/(n-1)}$$
A: Firstly; the answer is log(3). Intuituvely, just like the example you have mentioned, the fastest growing term in the summation is 3^n therefore, one can ignore the other terms and proceed by simplifying (1/n log(3^n)) which yields 3. To have a good justification for "ignoring" other terms, you may write the summation as;
3^n+3^(n/2)+...=3^n(1+3^(-n/2)+...) 
and then write: log(3^n(1+3^(-n/2)+...)) =log(3^n)+log(1+3^(-n/2)+...)) 
the second term is indeed equal to log(1) which is zero. Therefore you may proceed by calculating lim 1/nlog(3^n) which is solved by moving the exponential to beforehandn of the log. 
