Separating family of seminorms Let $(p_n)_{n=1}^{\infty}$ be a family of seminorms on a vector space $X$. Assume that series $\sum_{n=1}^{\infty} p_n(x)$ is convergent for any $x \in X$ and let's denote sum of such series as $q(x)$. 
I have to prove that $q: X\ni x\rightarrow q(x) \in [0,\infty)$ is a norm on $X$ if and only if $(p_n)_{n=1}^{\infty}$ is a separating family of seminorms on $X$.
Thanks for any help.
 A: $\implies$: Assume $q(x) = \sum_n p_n(x)$ defines a norm. You want to show that $p_n$ are seminorms and that they are separating, meaning that if $x\neq 0$ then there is $n$ such that $p_n(x) \neq 0$. First let's show $p_n$ are separating. Let $x \neq 0$. Then $q(x) \neq 0$. Then there must be at least one $p_n(x)$ not equal to zero. 
It remains to be shown that $p_n$ satisfiy the triangle inequality and that $cp_n(x) = p_n(cx)$. The latter follows immediately from $cq(x) = q(cx)$. For the triangle inequality assume by contradiction that there is $n$ and $x,y$ such that $p_n(x+y) > p_n(x) + p_n(y)$. Then $q(x+y) = \sum_n p_n(x+y) \ge \sum_n p_n(x) + \sum_n p_n(y) = q(x) + q(y)$, a contradiction. 
$\Longleftarrow$: If $p_n$ are seminorms then $q(x + y)\le q(x) + q(y)$ follows from the corresponding property of $p_n$. Similarly, $cq(x) = q(cx)$ is clear. To finish this direction you need to show that $q(x) = \sum_n p_n(x) = 0 $ implies $x=0$. This follows from the separating proprty: if $x$ was not zero then there would exist $p_n$ with $p_n(x) \neq 0$. But then we'd have $\sum_n p_n(x) \neq 0$.
