# Show that $\frac1{1^2}+\frac1{3^2}+\frac1{5^2}+\cdots =\frac{\pi^2}8$ if $\frac1{1^2}+\frac1{2^2}+\frac1{3^2} +\cdots =\frac{\pi^2}{6}$ [closed]

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.

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

Show that $$\frac{1}{1^2} +\frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}\cdots =\frac{\pi^2}{8}$$

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

## closed as off-topic by tired, Martin Sleziak, Leucippus, Daniel W. Farlow, suomynonAOct 26 '16 at 3:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – tired, Leucippus, Daniel W. Farlow, suomynonA
If this question can be reworded to fit the rules in the help center, please edit the question.

• Please create a more relevant title. – Jacob Oct 25 '16 at 15:52
• "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. – Namaste Oct 25 '16 at 16:04
• we had this hundreds of times here on this iste – tired Oct 25 '16 at 16:26
• @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. – Namaste Oct 25 '16 at 17:13
• 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. – Namaste Oct 25 '16 at 17:49

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.

• $+1.$ 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

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

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

• Still perplexed – Lanous Oct 25 '16 at 16:01
• @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
• @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). :) – tatan Oct 25 '16 at 16:14
• @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

If you are familiar with Euler products, you can see that

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

• lol, fancy you! :D – Paichu Oct 26 '16 at 1:34

\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}

• 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. – Namaste 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