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Problem:

If
$$\frac{1}{1^2} + \frac{1}{2^2} +\frac{1}{3^2} + \frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$$ show that $$\frac{1}{1^2} +\frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}\cdots =\frac{\pi^2}{8}$$

This was a question on an exam yesterday. My professor always throws one question in that is above our level and this was the one. I had no idea what to do on the exam. I just wanted to see an answer and the mode of thinking behind said answer.

I am an undergrad student. Sometimes it's helpful to see the work of others to understand a process. I'm an undergrad student in math. The class is history of mathematics.

I've tried many things, multiplying equations, subtracting equations. any help would be appreciated.

Lanous
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    Please create a more relevant title. – Jacob Oct 25 '16 at 15:52
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    "I've tried many things, multiplying equations, subtracting equations." Care to share any specifics of what you've tried? How would you feel if your only answer to this question was "Yeah, I solved/proved it, many times". Period, Nothing more. Wouldn't you want that answerer to expand on his/her answer and share more? Well, the flip side, you sharing what you've done, is just as important. – amWhy Oct 25 '16 at 16:04
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    we had this hundreds of times here on this iste – tired Oct 25 '16 at 16:26
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    The votes to close this as "off topic" are absurd. $\qquad$ – Michael Hardy Oct 25 '16 at 17:04
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    @MichaelHardy Have you not noticed all the questions that are closed each day for being [off-top/lack of context]? You know very well that the only reason people click on "off topic" is in order to identify the problem being the absence of any OP-provided context, which can be shown in many ways. This OP failed to address/add additional context, though directly asked upon the post of the question. Lack of context is one thing, and you should know, for me, it takes more than that; in this case, since the OP started receiving answers anyway, s/he decided s/he didn't have to work, anyway. – amWhy Oct 25 '16 at 17:13
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    @amWhy : It is absurd that "off-topic" is used for that purpose. There should be a separate menu item for that. But that proposal has been opposed vehemently, angrily, and boorishly. There seems to be a powerful clique devoted to vehement opposition to make more than the tiniest changes to anything about the status quo. Not only to oppose such things vehemently, but to treat disrespectfully those who propose such things. $\qquad$ – Michael Hardy Oct 25 '16 at 17:20
  • I agree that the context issue should be a first-page option. I've never ever opposed that, and I've always supported its independence from the "off topic" chain. At this point, if a post offers no context, and the op refuses to respond to requests for context after having been asked for it in comments, etc., I have little choice than to click on "off-topic" to be able to identify the problem as missing context. So don't point your finger at me. – amWhy Oct 25 '16 at 17:27
  • @MichaelHardy This comment exchange has been really off topic...or at least has reached a point where the conversation is less about the question at hand, and more about meta issues. I'm signing off. – amWhy Oct 25 '16 at 17:29
  • @amWhy this was a question on an exam yesterday. My professor always throws one question in that is above our level and this was the one. I just wanted to see an answer and the mode of thinking behind said answer. I am an undergrad student. Sometimes it's helpful to see the work of others to understand a process. This was all the context I was given on the test I took yesterday. – Lanous Oct 25 '16 at 17:40
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    That comment you just wrote would well serve as context. Providing context can happen in many forms, and what you just wrote in a comment, if edited within your question, would certainly served as sufficient context. Unless you tell us, we don't know if you're in the midst of similar material, or if it's your first "taste" of summations, etc. – amWhy Oct 25 '16 at 17:49
  • Not answering questions directly asked to the OP after an answer is worse than the really poor options htere are to close a post in this site...and much worse, and in several ways, is not to require anyone to write down some basic background in mathematics and to explain in short why a downvote. All this has come up tens of times in meta and everwhere else in the last 4- 5 years at least. Nothing will change, I'm afraid...not for the better, anyway. – DonAntonio Oct 25 '16 at 17:52
  • No, @Antonio, if you embrace fatalism, you stop trying, and, sure, self-fulfilling process, nothing will change. I have NEVER voted to close without some attempt on my part to help an OP obliterate any reason for closure. I can't pretend that my actions alone will be enough to save a question because other users may feel the need to act, or because an OP fails to act....BUT, that doesn't condone the belief and behavior that "Nothing will change, or "My actions in answering (problem statement questions bar all else), or commenting don't make any difference in the quality of questions." – amWhy Oct 25 '16 at 18:06
  • @DonAntonio When you start battling on behalf of OP's other than those to whom you've provided answers, everything you protest to is likely seen as self-interested. – amWhy Oct 25 '16 at 18:09
  • @amWhy I was more addressing Michael than you, but anyway: no condoning needed here: the "nothing will change here (for the better)" is just an observation, not even a judgement. No fatalism involved, though a good ammount of pesimism surely is. I agree with you that all this post, from the question itself and on, deserved to be closed at least as much as many others and, in fact, more than many that have been already closed. I just stressed my view that the OP no addressing questions directed to him is even worse than not showing own work and etc. – DonAntonio Oct 25 '16 at 18:17
  • KO @DonAntonio Thanks for the explanation. I get what you're saying. When I challenge others' "pessimism" I'm simultaneously trying to convince myself, too! – amWhy Oct 25 '16 at 18:21
  • Possible duplicate of how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? – Martin Sleziak Oct 25 '16 at 20:25

4 Answers4

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Hints: fill in details (explanations, justifications, etc.)

$$\frac{\pi^2}6=\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\cdot\frac{\pi^2}6+\sum_{n=1}^\infty\frac1{(2n-1)^2}$$

DonAntonio
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  • Still perplexed – Lanous Oct 25 '16 at 16:01
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    @Lanous What step exactly perplexes you ? Observe that the other answer (tatan's) repeats exactly the same as mine, just with lots of parentheses and dots... – DonAntonio Oct 25 '16 at 16:04
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    @DonAntonio Maybe, the OP cannot understand the summations. By the way, sorry for an absolute copy of your answer(Though I didn't mean to do so). :) – Soham Oct 25 '16 at 16:14
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    @tatan Don't worry. Perhaps yours is clearer to the OP. Anyway, somone messing with these series must, I believe, be able to read summations...and anyway: I asked him what's the problem and he didn't answer. – DonAntonio Oct 25 '16 at 16:16
  • @Lanous : Take a look at my answer. My claim would be that it's the simplest one posted so far. $\qquad$ – Michael Hardy Oct 25 '16 at 16:31
  • @DonAntonio your solution makes since after seeing tatan's. I'm taking a history of math class right now. I haven't had a lot of experience with these types of problems. So I am not always sure what I am allowed to do. I'm in my second to last semester from finishing my math degree and obviously have a lot left to learn. Thank you – Lanous Oct 25 '16 at 17:32
  • @Lanous Let $\zeta(s) = \sum_{n=1}^\infty n^{-s}$. Since $2^{-s} \zeta(s) = \sum_{n=1}^\infty (2n)^{-s}$ you have that $\zeta(s)(1-2^{-s}) = (\sum_{n=1}^\infty n^{-s})-(\sum_{n=1}^\infty (2n)^{-s}) = \sum_{n=1}^\infty (2n-1)^{-s}$. Here $s= 2$ – reuns Oct 25 '16 at 20:09
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Let the required sum be $S$.

$$\frac{1}{1^2} + \frac{1}{2^2} +\frac{1}{3^2} + \frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$$

$$\implies \frac{1}{1^2} +\frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}+\cdots +\frac{1}{2^2}+\frac{1}{4^2}+\cdots= \frac{\pi^2}{6}$$

$$\implies S+\frac{1}{2^2}\left(\frac{1}{1^2} + \frac{1}{2^2} +\frac{1}{3^2} + \frac{1}{4^2}+\cdots\right)= \frac{\pi^2}{6}$$

$$\implies S+\frac{1}{4} \cdot \frac{\pi^2}{6}=\frac{\pi^2}{6}$$

$$\implies S=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$$

Hope this helps.

Michael Hardy
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Soham
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  • $+1.$ This is pretty good, but as the local worshipper of simplicity, I've attempted to make it simpler. See my answer. $\qquad$ – Michael Hardy Oct 25 '16 at 16:30
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If you are familiar with Euler products, you can see that

$$1+{1\over3^2}+{1\over5^2}+{1\over7^2}+\cdots=\prod_{p\not=2}\left(1-{1\over p^2}\right)^{-1}=\left(1-{1\over 2^2}\right)\prod_p\left(1-{1\over p^2}\right)^{-1}={3\over4}\left(1+{1\over2^2}+{1\over3^3}+{1\over4^2}+\cdots\right)$$

Barry Cipra
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\begin{align} & \frac 1 {1^2} + \frac 1 {3^2} + \frac 1 {5^2} + \frac 1 {7^2} + \cdots \\[10pt] = {} & \left( \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^3} + \frac 1 {4^2} + \frac 1 {5^2 } + \frac 1 {6^2} + \cdots \right) - \left( \frac 1 {2^2} + \frac 1 {4^2} + \frac 1 {6^2} + \cdots \right) \\[10pt] = {} & \left( \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^3} + \frac 1 {4^2} + \frac 1 {5^2 } + \frac 1 {6^2} + \cdots \right) - \frac 1 {2^2} \left( \frac 1 {1^2} + \frac 1 {2^2} + \frac 1 {3^2} + \cdots \right) \\[10pt] = {} & \frac {\pi^2} 6 - \frac 1 4\cdot\frac{\pi^2} 6. \end{align}

Michael Hardy
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  • This just seems to be an algebraic manipulation to @tatan's answer, so I'm not clear why you posted it roughly 1/2 hour after the others? Not saying your answer is incorrect; rather, redundant. – amWhy Oct 25 '16 at 16:35
  • @amWhy : I think things should be made as simple as they really are. That answer has some arrows ($\Longrightarrow$) and a variable that's not just a Hindu‒Arabic numeral. $\qquad$ – Michael Hardy Oct 25 '16 at 16:44
  • It is my opinion that a good student, after (s)he meets an answer, and it can be mine, tatan's or this one, must be able, at this level, to break it apart in parts and realize the three are just the very same. A little more interesting is to ask why can we split things that way (absolute convergence and etc.), but it's up to the OP to check and, in case of need, to ask for further explanation. Barry's answer is very nice but likely over the apparent level of the OP. I don't think things "should" be made as simple as possible: let the asker do that after (s)he already understood the gist. – DonAntonio Oct 25 '16 at 16:50
  • @DonAntonio : I disagree. Of course a good student would see that these are in a sense all the same, but that reason doesn't impress me. Making things as simple as the really are takes work, and is beyond what even a good student, or most mathematicians, can usually do. $\qquad$ – Michael Hardy Oct 25 '16 at 16:52
  • $\ldots,$and very often I am appalled by how messy people make answers when they could be expressed far more simply. They flunked poetry, and poetry is important. $\qquad$ – Michael Hardy Oct 25 '16 at 16:54
  • @MichaelHardy Maybe you've a point about simplicity, yet I disagree the answerer "should" do that...and I also not that sure yours is simpler than tatan's...or for that matter and for somebody that really understands, than mine. And I think that, in general, poetry sucks as much as general literature in prose rocks. – DonAntonio Oct 25 '16 at 16:54
  • Doing things in a way that's suitable "for anybody that really understands" is a mistake. $\qquad$ – Michael Hardy Oct 25 '16 at 16:56
  • Poetry is also necessary for understanding. $\qquad$ – Michael Hardy Oct 25 '16 at 17:03
  • @MichaelHardy "For anybody that understands"... the basic notions, definitions and facts of what he asks about and/or after reading the (one) answer. Don't take my statement out of its context: I am also a lecturer and, imo, a really darn good one. That's the reason that is not a mistake, at least not in this site, where one has to guess what people know and what they don't. And nop: I really think no poetry at all is necessary to understand, at least mathematics. Perhaps other things... – DonAntonio Oct 25 '16 at 17:47