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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`.2012-05-14
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    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 $\alpha\to0$?2012-05-14
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    Either that or $\sum\limits_{i=1}^m\frac{w_i}{n_i}=m$.2012-05-14
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    Note that my comment related to an earlier version of the problem. But there is still no reply to Antonio's questions.2012-05-14
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    @GerryMyerson It is one step in a financial essay.2012-05-14
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    @GerryMyerson I revise the formula2012-05-14
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    What happens if you try l'Hopital's Rule?2012-05-14
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    @GerryMyerson can I use it?2012-05-14
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    Well, you have the indeterminate form $0/0$, so I don't see why you can't use it. I haven't tried it myself, so I don't guarantee it will work, but it seems to me it's worth a try.2012-05-14

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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.