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

  • 3
    $\begingroup$ The sum is actually $\pi^2/8$, according to Wolfram Alpha. $\endgroup$ – Rahul Aug 21 '16 at 19:45
  • 3
    $\begingroup$ 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 $\endgroup$ – Winther Aug 21 '16 at 23:21
  • $\begingroup$ $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2s}} = \sum_{n=1}^{\infty}\frac{1}{n^{2s}}-\sum_{n=1}^{\infty}\frac{1}{(2n)^{2s}} = (1-\frac{1}{2^{2s}})\sum_{n=1}^{\infty}\frac{1}{n^{2s}} = (1-\frac{1}{2^{2s}})\zeta(2s) = (1-\frac{1}{2^{2s}})\frac{(2\pi)^{2s}|B_{2s}|}{2(2s)!}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values $\endgroup$ – Beauty Is Truth Aug 21 '16 at 23:22
  • $\begingroup$ For positive $a$ and $b,$ $$\sqrt a+\sqrt b=\sqrt{a+b}\iff ab=0.$$ $\endgroup$ – Bumblebee Nov 7 '16 at 19:17

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.

  • 6
    $\begingroup$ 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. $\endgroup$ – fleablood Aug 21 '16 at 22:24
  • 3
    $\begingroup$ 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 $\endgroup$ – P Vanchinathan Aug 22 '16 at 0:22
  • 5
    $\begingroup$ 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. $\endgroup$ – fleablood Aug 22 '16 at 0:30
  • 1
    $\begingroup$ I refuse to get rid of the "more". "more helpy" stays. $\endgroup$ – fleablood Aug 22 '16 at 18:32
  • 1
    $\begingroup$ 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. $\endgroup$ – fleablood Aug 22 '16 at 20:25

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.

  • 1
    $\begingroup$ 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). $\endgroup$ – Lubin Aug 22 '16 at 17:55

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!)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.