$\lim_{p\to 0^+}\left(\sum_{n=1}^N\lambda_na_n^p)\right)^{1/p}=\exp\left(\sum_{n=1}^N\lambda_n\log(a_n)\right)$

Question

I have to check that $$\lim_{p\to 0^+}\left(\sum_{n=1}^N\lambda_na_n^p)\right)^{1/p}=\exp\left(\sum_{n=1}^N\lambda_n\log(a_n)\right),$$ where $a_i>0$, for all $i=1,\ldots , N$ and $\sum_{n=1}^N \lambda_n=1$. Can anyone help me?

Answer 1

Say $b_j=\log(a_j)$. Then $$a_j^p=e^{pb_j}=1+pb_j+O(p^2),$$so $$\sum\lambda_ja_j^p=1+p\sum\lambda_jb_j+O(p^2).$$ Since $\log(1+t)=t+O(t^2)$ as $t\to0$ this shows that $$\log\left(\sum\lambda_ja_j^p\right)=p\sum\lambda_j b_j+O(p^2),$$ so $$\frac1p\log\left(\sum\lambda_ja_j^p\right)=\sum\lambda_j b_j+O(p)\to\sum\lambda_jb_j\quad(p\to0^+).$$