Can I Square Root A Series? I have a question in Quantum Mechanics where I need to solve a series, and the thing is that I can get the answer to a similar series with the help of the same problem but I am not sure if I can square root my series to use it in the problem. For example I have $$\sum_{n=1,3,5,7...}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{96}$$. But the summation I need is for $\style{Bold}{\frac{1}{n^2}}$ So is it fine to square root both sides and say $$\sum_{n=1,3,5,7...}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{\sqrt{96}}$$
 A: 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. 
A: Yes, series can be square rooted, but not like this. Try squaring a series and then finding what it is when you make it equal to original series. It wont be pretty but it will be the square root of series.
A: The already given answers are quite nice; here is one more way to think about this:
The question as to why the answer to your question is no (or, at least, "not as simply as you might wish") is interesting. But how do you establish, in the first place, that the answer is no?
To do so, it will suffice to come up with a single counterexample. Picking counterexamples is a bit of an art; for the question proposed here, I am inclined to try an infinite series for which the sum converges to $1$. The reasoning in my mind is, I can then check whether the corresponding square-rooted series converges to $\sqrt{1} = 1$. (Or is it $\sqrt{1} = -1$? Well, let us investigate.)
Consider the classic geometric series:
$$\sum_{n = 1}^{\infty} \frac{1}{2^n} = 1/2 + 1/4 + 1/8 + \cdots = 1$$
Next, we consider the square-rooted version:
$$\sum_{n = 1}^{\infty} \frac{1}{2^{n/2}} = 1/\sqrt{2} + 1/\sqrt{4} + \cdots $$
The latter series has all positive terms, so its partial sums are monotonically increasing. Yet we add up just the first two terms to find:
$$1/\sqrt{2} + 1/\sqrt{4} = 1/\sqrt{2} + 1/2 > 1$$
So, whatever is going on with the latter series (which, incidentally, does converge: to $1 + \sqrt{2}$) we know its computation is not as simple as taking the square root of our first series' limit; that simplistic approach would suggest the square-rooted series would also converge to $1$, but already two terms in it has surpassed $1$ with no plan to return.
(The accepted answer provides an even more extreme example: An initial series, summing the squares' reciprocals, which converges, but a square-rooted version that yields the harmonic series, which diverges!)
A: You are asking if $\sqrt{(\sum a_i)} = \sum(\sqrt {a_i}) $.
Does it?  Does $5=\sqrt {9 +16} = \sqrt {9} + \sqrt {16}=7 $?
A: 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.
