Sum of probability differential is zero

Question

I keep seeing:

$$ \sum_i [ln(p_i) + 1]dp_i = \sum_i ln(p_i)dp_i $$

where $p_i$ is the probability of the $i^{th}$ state and where

$ \sum_i p_i = 1 $

cropping up in my statistical mechanics course. I'm a bit unsure of how one comes to this?

EDIT: A similar result which confuses me is:

$$ \sum_i [ln(Z)] dp_i = 0 $$

where $Z$ is the partition function.

Answer 1

Presumably because we always have $ \sum_i p_i = 1$: if we differentiate this condition, $$ \sum_i dp_i = 0 $$ so that the total probability remains $1$. It's like $\dot{x} \cdot x = 0 $ when $x$ is forced to lie on a sphere $x \cdot x = R^2$.