Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $y_i$

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.