Can I Square Root A Series?

No. For the same reason that $$ \sqrt{a_1+...+a_n}\ne \sqrt{a_1}+....+\sqrt{a_n} $$ What you are saying would also imply that say the harmonic series converges, which it does not.

You are asking if $\sqrt{(\sum a_i)} = \sum(\sqrt {a_i}) $.

Does it? Does $5=\sqrt {9 +16} = \sqrt {9} + \sqrt {16}=7 $?

Less rude, more helpy:

There is the Schwarz inequality:

$(\sum ab)^2 \le \sum a^2 * \sum b^2$

So $\sum 1/n^4 = \pi^4/96$ obviously does not mean $\sum 1/n^2 = \pi^2/\sqrt{96}$ but it does mean $\sum 1/n^2 \ge \pi^2/\sqrt{96}$.

It further means $\sum 1/n \ge \pi/\sqrt[4]{96}$ which it is.

$\sqrt{a} + \sqrt{b} \ge \sqrt{a + b}$ (hence $7 = \sqrt{9} + \sqrt{16} \ge \sqrt{9+16} = 5$.

That can be useful.