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What kind of algebraic manipulations are required to show the following?

$$ \frac2{1^2}-\frac2{2^2}+\frac2{3^2}-\frac2{4^2}+\cdots=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots. $$

The LHS has alternating signs and is twice the RHS, and the series converges to $\pi^2/6$.

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By the way: – Argon Oct 28 '12 at 17:27
I think you want to multiply your LHS by $-1$. – Benjamin Dickman Oct 28 '12 at 17:27
up vote 3 down vote accepted

You are missing a negative sign. The series sums to $-\pi^2/6$. Consider $$S = \dfrac1{1^2}-\dfrac1{2^2}+\dfrac1{3^2}-\dfrac1{4^2} \pm $$ The series you are interested in is nothing but $-2S$. $$S = \dfrac1{1^2}+\dfrac{1-2}{2^2}+\dfrac1{3^2}+\dfrac{1-2}{4^2} + $$ $$S = \underbrace{\left(\dfrac1{1^2}+\dfrac{1}{2^2}+\dfrac1{3^2}+\dfrac{1}{4^2} + \right) - 2 \left(\dfrac1{2^2}+\dfrac{1}{4^2}+\dfrac1{6^2}+\dfrac{1}{8^2} + \right)}_{\text{Allowed to rearrange since }S \text{ is absolutely convergent}} = \zeta(2) - \dfrac2{2^2} \zeta(2) = \dfrac{\zeta(2)}{2}$$ Hence, the answer is $-2S = - \zeta(2)$.

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Since both sides converge, we can interleave terms, so since

$$ RHS = \frac4{2^2}+\frac4{4^2}+\frac4{6^2}+\frac4{8^2}+\cdots$$

we have

$$ RHS - LHS = \frac2{1^2}+\frac2{2^2}+\frac2{3^2}+\frac2{4^2}+\cdots = 2 RHS$$

so $$RHS= - LHS$$

proving the equality but with a change of sign. Indeed, by grouping pairs of terms, the left hand side is clearly negative.

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move the negatives to the other side


add 2x odd terms to get all the 3s




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Perhaps I'm missing something really elementary here, but isn't your last equality actually saying that $$\sum_{n=1}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{3}{n^2}...??$$ which, of course, would be wrong.... – DonAntonio Oct 28 '12 at 17:48
The OP has corrected the sign error in the original post, so you might want to amend your answer to reflect that (and end up with a correct statement), or just delete your answer. – Cameron Buie Oct 28 '12 at 18:36

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