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?

Hint. $$0<3^{\frac n 2}+\cdots+3^{\frac n n}\le(n-1)3^{\frac n 2}=\frac{3^n}{3^{\frac n 2}/(n-1)}$$
