I'm having trouble understanding the solution of a limit. It involves a formula for measuring certainty of a discrete probability distribution.
Given a set of values $p_j$ which sum up to 1, find the limit $\lim_{\alpha\to 1} (\sum p_j^\alpha)^\frac{1}{\alpha-1}$. The author's solution is as follows:
1) Take the natural logarithm, yielding $\lim_{\alpha\to 1} \frac{1}{\alpha-1}\ln\sum p_j^\alpha$.
2) Take the derivative with respect to $\alpha$ of the numerator and denominator to get $\lim_{\alpha\to 1}\frac{\ln\sum p_j^\alpha}{\alpha-1} = \lim_{\alpha\to 1}\frac{\sum p_j^\alpha \ln p_j}{\sum p_j^\alpha}$.
3) We can then plug in $\alpha$, and since $\sum p_j = 1$, we can reduce to $\sum p_j \ln p_j$.
4) Then we undo the natural log to get $\prod p_j^{p_j}$.
I verified the correctness of the result with a few values on WolframAlpha.
The step I don't understand is step 2. When I try to take the derivative of the numerator, I use the chain rule and get $\frac{d}{d\alpha}\ln\sum p_j^\alpha=\frac{1}{\sum p_j^\alpha}\times\sum\alpha p_j^{\alpha-1}$. This simplifies to give an answer larger than 1, which makes no sense.
How do you get $\frac{d}{d\alpha}\ln\sum p_j^\alpha = \frac{\sum p_j^\alpha \ln p_j}{\sum p_j^\alpha}$?