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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 $$\sum_{n=1,3,5,7...}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{\sqrt{96}}$$

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1  
The sum is actually $\pi^2/8$, according to Wolfram Alpha. – Rahul 13 hours ago
1  
As for the sum you are after: we have $\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=2,4,6,\ldots}^\infty \frac{1}{n^2} +\sum_{n=1,3,5,\ldots}^\infty \frac{1}{n^2} = \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} +\sum_{n=1,3,5,\ldots}^\infty \frac{1}{n^2}$ so the sum is $\sum_{n=1,3,5,\ldots}\frac{1}{n^2} = \frac{3}{4} \sum_{n=1}^\infty \frac{1}{n^2} = \frac{3}{4}\frac{\pi^2}{6} = \frac{\pi^2}{8}$. How to derive the famous result $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ is found here – Winther 10 hours ago
    
$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2s}} = \sum_{n=1}^{\infty}\frac{1}{n^{2s}}-\sum_{n=1}^{\infty}\frac{1}{(2n)^{2s}} = (1-\frac{1}{2^{2s}})\sum_{n=1}^{\infty}\frac{1}{n^{2s}} = (1-\frac{1}{2^{2s}})\zeta(2s) = (1-\frac{1}{2^{2s}})\frac{(2\pi)^{2s}|B_{2s}|}{2(2s)!}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values – Beauty Is Truth 10 hours ago
up vote 20 down vote accepted

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.

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up vote 5 down vote

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.

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Well, in all honesty, you asked an extremely obvious question that you really should have learned the answer to in middle school Still we shouldn't be rude and we should still try to help. I figure you might not know the schwarz inequality which migh be helpful for what you are trying to do. – fleablood 11 hours ago
1  
Thus comment by Schrodinger is uncalled for as it us unprofessional and makes this personal. Fleablood has been professional and shown ecemplary politeness. I appreciate his restraint – P Vanchinathan 9 hours ago
3  
Actually, I think I have been averagely rude. This began with a now deleted comment that was somewhat rude but easy to understand comment that shrodinger should review his calculus notes. Shrodinger got huffy at this and then it was pointed out rather bluntly that this is a very basic at a pre-algebra level. I got a little rude but I felt bad about it. Ultimately we are here to answer questions. – fleablood 9 hours ago
up vote 4 down vote

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

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

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up vote 2 down vote

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.

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