Let $m \in \mathbb{N}$ be fixed and $\forall i \in \{1,2,...m\}, a_{i}\geq0$ Now, define $S_{n} = \bigg( \sum_{i=1}^{m}\frac{1}{m}a_{i}^{\frac{1}{n}}\bigg)^{n}$. I want to show that $\lim\limits_{n\to\infty}S_{n} = \frac{1}{m} \sum_{i=1}^{m}a_{i}$. In the other words, $S_{n}$ converges to arithmetic mean.
So, my attempt so far is to to apply sandwich theorem. By using Jensen's inequality, I can obtain $$0\leq S_n \leq \frac{1}{m}\sum_{i=1}^{m}a_{i}$$ My problem is the lower bound since I can only obtain $\frac{1}{m^{n}}\sum_{i=1}^{m}a_{i}$ which tends to $0$ as $n\to\infty$.
Any help to obtain the lower bound or any hint to use another method might be nice.
Thank you very much!
We suppose that $a_i\gt 0$ for all $i$. If not, the limit is zero because in this case, $$ S_n\leqslant \left(\frac{m-1}m\right)^n\max_{1\leqslant i\leqslant m}a_i. $$
First observe that $$ \ln S_n=n\ln\left(1+\frac 1m\sum_{i=1}^m\left(a_i^{1/n}-1\right)\right). $$ Letting $b_n:= \frac 1m\sum_{i=1}^m\left(a_i^{1/n}-1\right)$, we rewrite this as $$ \tag{1}\ln S_n=nb_n\frac{\ln\left(1+b_n\right)}{b_n}. $$ Since $\lim_{t\to 0}\frac{\ln (1+t)}t=1$ and $b_n\to 0$, it suffices to look at the convergence of $nb_n$. To this aim, show that for all positive number $x$, $n\left(x^{1/n}-1\right)\to \ln x$ (use the derivative of exponential function for example). It follows that $$ \tag{2}\lim_{n\to +\infty}nb_n=\frac 1m\sum_{i=1}^m\ln a_i. $$ The combination of (1), (2) and continuity of the $\exp$ function gives the result.