Try to evaluate $\lim_{\alpha\to1^-}\frac{1}{\alpha(\alpha-1)}\left[\frac{1}{m}\sum_{i=1}^{m}\left(\frac{y_i}{y^{'}}\right)^\alpha-1\right]$

Question

I can't evaluate this limit. $$\lim_{\alpha\to1^-}\frac{1}{\alpha(\alpha-1)}\left[\frac{1}{m}\sum_{i=1}^{m}\left(\frac{y_i}{y^{'}}\right)^\alpha-1\right]$$ where $y_i>0$, $y^{'}$ is the average of $y_i$

Answer 1

Let $$ \bar{y}=\frac1m\sum_{i=1}^my_i\quad\text{and}\quad x_i=\frac{y_i}{\bar{y}} $$ Then $$ \begin{align} \lim_{\alpha\to1^-}\frac{1}{\alpha(\alpha-1)}\left[\frac{1}{m}\sum_{i=1}^{m}\left(\frac{y_i}{\bar{y}}\right)^\alpha-1\right] &=\lim_{\alpha\to1^-}\frac{1}{\alpha-1}\left[\frac{1}{m}\sum_{i=1}^{m}x_i^\alpha-1\right]\\ &=\lim_{\alpha\to1^-}\frac{1}{\alpha-1}\left[\frac{1}{m}\sum_{i=1}^{m}\left(x_i^\alpha-x_i\right)\right]\\ &=\frac1m\sum_{i=1}^mx_i\log(x_i)\\ &=\frac1m\sum_{i=1}^m\frac{y_i}{\bar{y}}(\log(y_i)-\log(\bar{y}))\\ &=\frac{{\small\displaystyle\sum_{i=1}^m}\;y_i\log(y_i)}{{\small\displaystyle\sum_{i=1}^m}\;y_i}-\log(\bar{y}) \end{align} $$ I don't see why this wouldn't work for $\alpha\to1^+$, too.