what is summation in the sense of a principal value? In one  paper I  saw  this equality:
$$\sum_{\eta=-\infty}^{\infty}\frac{z}{(z+\eta)}=\pi z\cot(\pi z)$$
which is the same as
$$\sum_{\eta=-\infty}^{\infty}\frac{1}{(z+\eta)}=\pi \cot(\pi z)$$
where summation is understood in the sense of a principal value. What does it mean?
In another paper I found the next expression:
$$\frac{\exp(2\pi iaz)}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z-n}$$
for $a=0$ it  is equivalent  to
$$\frac{1}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\frac{1}{z+n}$$
which is not exactly the same expression like in the first case.
$$\sum_{n=-\infty}^{\infty}\frac{1}{z+n}=\pi Cot[\pi z]-i\pi$$
Where is my mistake? 
If the second formula is wrong, what is the correct formula for the second case?
$$\sum_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z+n}=?$$
 A: A principal-value sum (or integral) is usually one in which unconditional summation (or integration) does not converge, so one needs to sum in a particular way to achieve convergence.  I suspect that, in this case, the necessary summation is symmetric, so that we consider $\lim_{N \to \infty} \sum_{n = -N}^{n = N} f(n)$ instead of $\sum_{n = 1}^\infty f(-n) + \sum_{n = 0}^\infty f(n)$.
It's not quite clear to me what your issue is with the two formulæ you mention.  Since you are summing different functions ($1/(z + n)$ versus $z/(z + n)$), it is no surprise that the answers are different.  What am I missing?  (Sorry, I did not notice that you had already factored out the $z$.)
A: See the answer of L Spice for the principal value bit.
For the second bit, the formula from the second paper is rather suspect. For example, $a=1/2$ produces the divergent sum (even in the principal value sense) $$\sum_{n=-\infty}^\infty \frac{(-1)^n}{z-n}.$$ And for your case $a=0$, $z=1/2$ yields $\sum(z-n)^{-1}=0$ by symmetry, so the formula cannot be right then either.
