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

Caleb Stanford
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Shrodinger 2016
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    The sum is actually $\pi^2/8$, according to Wolfram Alpha. –  Aug 21 '16 at 19:45
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    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 Aug 21 '16 at 23:21
  • $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2s}} = \sum_{n=1}^{\infty}\frac{1}{n^{2s}}-\sum_{n=1}^{\infty}\frac{1}{(2n)^{2s}} = (1-\frac{1}{2^{2s}})\sum_{n=1}^{\infty}\frac{1}{n^{2s}} = (1-\frac{1}{2^{2s}})\zeta(2s) = (1-\frac{1}{2^{2s}})\frac{(2\pi)^{2s}|B_{2s}|}{2(2s)!}$ https://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values – Beauty Is Truth Aug 21 '16 at 23:22
  • For positive $a$ and $b,$ $$\sqrt a+\sqrt b=\sqrt{a+b}\iff ab=0.$$ – Bumblebee Nov 07 '16 at 19:17

5 Answers5

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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.

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

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

fleablood
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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.

fleablood
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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 Aug 21 '16 at 22:24
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    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 Aug 22 '16 at 00:22
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    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 Aug 22 '16 at 00:30
  • To change the subject, I think the comparative of “helpy” is “helpier”. – Lubin Aug 22 '16 at 17:43
  • "more helpier" is redundant. – fleablood Aug 22 '16 at 17:48
  • I didn’t recommend “more helpier”, just “helpier”. – Lubin Aug 22 '16 at 17:50
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    I refuse to get rid of the "more". "more helpy" stays. – fleablood Aug 22 '16 at 18:32
  • I don't mean to be rude, but I am not necessarily convinced that your first revision of this post (containing only the first three sentences of this answer) are more helpful than the other answer(s), nor am I sure that it is less rude than the other answer(s). To be honest, it is not clear at all whether your first remark is just stating something about all other answers or just about your own other answer. If it is the former, than I would consider this answer to be more rude (to other answerers) than your other answer is to the OP. – wythagoras Aug 22 '16 at 19:42
  • I would further like to note that I am posting this and the previous comment while not seeing any deleted comment(s) on this or the other answer. – wythagoras Aug 22 '16 at 19:45
  • Oh, I'm not claiming I am being rude-free. I am pointing out $\sqrt {sum x} \ne \sum \sqrt{x}$ and I'm slightly rudely implying that that's a dope-slappingly stupid mistake to make. However the schwarz inequality that $(\sum ab)^2 \ge \sum a^2 \sum b^2$ is more information that can be made useful if properly applied and was thus useful information. The comment "less rude; more helpy" was definitely in response to the my other answer. The other answers aren't in the least bit rude. The deleted comments were... – fleablood Aug 22 '16 at 20:13
  • ...i) a comment that if this is about quantuum mechanics the op presumable took calculus and should review notes. That's more snarky than rude. ii) It was replied with a comment that that was rude and unless everyone got A+ at MIT levels with perfect recall it was rude to complain. iii) in turn that got a omment that this wasn't A+ level MIT but very basic result and he really should review notes, getting v) a very huffy I'll leave math to the mathematicians but give me the results leading to my very blunt vi) This isn't a calculus so much as a middle school pre-algebra result. – fleablood Aug 22 '16 at 20:19
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    Basically the "rudeness" is about answering "So is it fine to square root both sides (of a sum)" with "Of course not! Don't be an idiot!" which is, admittedly, pretty rude. – fleablood Aug 22 '16 at 20:25
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    not to be rude, but I find the answer with the example helpier. Why? I can verify that while being half asleep. Plus, it teaches creativity and a method to quickly disprove a claim by finding just one counter-example. – atmelino Aug 22 '16 at 22:41
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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.

jimjim
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    Indeed, the square root of $1+a_1x+a_2x^2+a_3x^3 +a_4x^4+\cdots$ is $1+\frac12a_1x+(\frac12a_2-\frac18a_1^2)x^2+(\frac12a_3 - \frac14a_1a_2 + \frac1{16}a_1^3)x^3+$ (I don’t even want to write out the $x^4$-term). – Lubin Aug 22 '16 at 17:55
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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!)

Benjamin Dickman
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