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Prove that $$(\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots)\cdot(\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots)=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+\cdots$$

The product is derived from the summation:

$$\dfrac{\pi}{2\sqrt{2}}=\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots$$ Which upon squaring results in $\dfrac{\pi^2}{8}$

Use the identity: $$1=(1+\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})(1+\frac{1}{9})(1+\frac{1}{11})\cdots$$

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  • $\begingroup$ Is it necessary to use the identity 1=....? Because it can be easily proved without it $\endgroup$
    – Aarush Saharan
    Commented Sep 29, 2023 at 5:09
  • $\begingroup$ What are your thoughts? What approaches have you tried? $\endgroup$
    – Tony Mathew
    Commented Sep 29, 2023 at 5:13
  • $\begingroup$ @AarushSaharan not necessarily, I thought it would make the proof easier since it looked similar. By all means try to prove it however you'd like :) $\endgroup$
    – user1211726
    Commented Sep 29, 2023 at 5:27
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    $\begingroup$ @TonyMathew I've tried to prove it using the formula: $\sum{\dfrac{1}{r_i\cdot r_j}}=a_2/a_0$ for a polynomial. But I assumed sine function as the polynomial. This is kind of true(The result is 0 since sine function has no $a_2$ coefficient) but my goal is to prove the relation without any references to trigonometry, since the sine factorisation already proves the Basel problem. $\endgroup$
    – user1211726
    Commented Sep 29, 2023 at 5:30

2 Answers 2

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Now since you know that expansion for $\frac{\pi}{2\sqrt2}$ and as squaring it would give us

$(\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots)\cdot(\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots)=\dfrac{\pi^2}{8} $

also we know that

$ \dfrac{\pi^2}{6}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}+\dfrac{1}{10^2}+\cdots $

Now consider all the terms in the previous expansion with even denominators and take $\dfrac{1}{2^2}$ common

Hence we get

$ \dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+\cdots +\dfrac{1}{2^2}(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}+\dfrac{1}{10^2}+\cdots)=\dfrac{\pi^2}{6} $

The term with 0.25 common is also $\dfrac{\pi^2}{6}$ and taking it on the other side and subtracting you will get that the sum of squares of reciprocals of all odd numbers is also $\dfrac{\pi^2}{8}$

And so the result follows

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expanding $(\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots)\cdot(\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\cdots)$

and writing in the form of a matrix gives us:

$\begin{bmatrix}1 & +\frac{1}{3} & -\frac{1}{5} & -\frac{1}{7}& +\frac{1}{9}& +\frac{1}{11}& -\frac{1}{13}& -\frac{1}{15}& +\frac{1}{17} &+\cdots\\ \frac{1}{3} & +\frac{1}{3^2} & -\frac{1}{15}& -\frac{1}{21}& +\frac{1}{27}& +\frac{1}{33}& -\frac{1}{39}& -\frac{1}{45} & +\frac{1}{51} & +\cdots \\ -\frac{1}{5} & -\frac{1}{15} & +\frac{1}{5^2}& +\frac{1}{35}& -\frac{1}{45}& -\frac{1}{55}& +\frac{1}{65}& +\frac{1}{75} & -\frac{1}{85} & +\cdots\\ -\frac{1}{7} & -\frac{1}{21} & +\frac{1}{35}& +\frac{1}{7^2}& -\frac{1}{63}& -\frac{1}{77}& +\frac{1}{91}& +\frac{1}{105} & -\frac{1}{119} & +\cdots\\ \frac{1}{9} & +\frac{1}{27} & -\frac{1}{45}& -\frac{1}{63}& +\frac{1}{9^2}& +\frac{1}{99}& -\frac{1}{117}& -\frac{1}{135} & +\frac{1}{153} & +\cdots\\ \cdots \end{bmatrix}$

The diagonal is what we want.

The sum of the upper half above the diagonal = $(1+\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})(1+\frac{1}{9})(1+\frac{1}{11})\cdots - 1 = 0$

Similarly the sum of the bottom half below the diagonal = $(1+\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})(1+\frac{1}{9})(1+\frac{1}{11})\cdots - 1 = 0$

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