# Show that $\lim_{n\to\infty}\sqrt[\sum_\limits{k=0}^{n}\lambda_k]{\prod_\limits{k=0}^{n}a_k^{\lambda_k} }= \ell =\lim_{n\to\infty} a_n$

Suppose that $a_n\to \ell\neq 0$ is a converging sequence of positive real numbers and $\{\lambda_n\}$ is a sequence of positive real numbers such that $\sum\limits_{k=0}^{\infty}\lambda_k = \infty$.

Then show that, $$\lim_{n\to\infty}\sqrt[\sum_\limits{k=0}^{n}\lambda_k]{\prod_\limits{k=0}^{n}a_k^{\lambda_k} }= \ell =\lim_{n\to\infty} a_n$$

I have no clue on how to start.

The question is equivalent to:$$\lim_{n\to\infty}\frac{{\sum\limits_{k = 0}^n {{\lambda _k}\ln ({a_k})} }}{{\sum\limits_{k = 0}^n {{\lambda _k}} }} = \ln l$$ Since ${\sum\limits_{k = 0}^n {{\lambda _k}} }$ increases to $+\infty$, Stolz's theorem immediately gives the reult.
from this General Cesaro $\lim\limits_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k}\sum_\limits{k=0}^{n}\lambda_k a_k =\lim\limits_{n\to\infty} a_n$ we have $$\lim\limits_{n\to\infty} \frac{1}{\sum_\limits{k=0}^{n}\lambda_k}\sum_\limits{k=0}^{n}\lambda_k \ln(a_k) =\lim\limits_{n\to\infty} \ln(a_n)$$ then the result follows by taking to the exponential.