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

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 $y_i$

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Please use \lim and \sum. – Did May 14 '12 at 5:42
If $w_1=n_2=1/3$ and $w_2=n_1=2/3$, then the conditions are met, the sum is $1-(1/2)^{\alpha}+1-2^{\alpha}$ which goes to $-1/2$ as $\alpha\to1$, and the limit doesn't exist. Maybe it should be $\alpha\to0$? – Gerry Myerson May 14 '12 at 5:47
Either that or $\sum\limits_{i=1}^m\frac{w_i}{n_i}=m$. – anon May 14 '12 at 5:57
Note that my comment related to an earlier version of the problem. But there is still no reply to Antonio's questions. – Gerry Myerson May 14 '12 at 7:17
@GerryMyerson can I use it? – 89085731 May 14 '12 at 7:44

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.

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