Given a sequence of signed measures $<\nu_j>$, if it happens that $\nu=\sum\limits_{j = 1}^\infty \nu_j$ is still a valid signed measure (then it can be proved that each partial sum $\nu_n=\sum\limits_{j = 1}^n \nu_j$ is valid signed measure), do we have $\lim\limits_{n\to \infty}\sum\limits_{j = 1}^n \nu_j=\sum\limits_{j = 1}^\infty \nu_j$ ($=\lim\limits_{n\to \infty} \sum\limits_{j = 1}^n \nu_j$)? Thanks!

Assuming I understood the question correctly, the answer is no. Consider measures on $\{0,1\}^\omega$ with the product topology and Borel $\sigma$algebra. Let $\mu_i$ be the uniform measure on the set with $i$th coordinate equal to 0. This sequence converges by your definition to the uniform measure, but all $\mu_i$ are far (in total variation) from the uniform measure. (To fit your description we can take $\nu_i=\mu_i\mu_{i1}$). 


Am I misunderstanding your question? Because it seems to me that if you consider the $\nu_n = \frac{(1)^n}{n} \delta$ where $\delta$ is the Dirac mass, then we've got the usual situation with the alternating harmonic series. 

