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Let $(a_n)$ with $a_n \in \mathbb{R}$ be such that

$$\sum_{k = 0}^{\infty}\sum_{r = 0}^{2k}\Bigl|\frac{(-1)^k(2\pi)^{2k} a_r}{(2k - r + 1)!}\Bigr|< \infty$$

Why is following wrong? (Or is it correct?)

$$\sum_{k = 0}^{\infty}\sum_{r = 0}^{2k}\frac{(-1)^k(2\pi)^{2k} a_r}{(2k - r + 1)!} = \sum_{r = 0}^{\infty}\sum_{2k \geq r}^{}\frac{(-1)^k(2\pi)^{2k} a_r}{(2k - r + 1)!}$$

$$ = \sum_{r = 0}^{\infty}\sum_{2k \geq 2r}^{}\frac{(-1)^k(2\pi)^{2k} a_{2r}}{(2k - 2r + 1)!} + \sum_{r = 0}^{\infty}\sum_{2k \geq 2r + 1}^{}\frac{(-1)^k(2\pi)^{2k} a_{2r + 1}}{(2k - 2r)!}$$

$$ = \sum_{r = 0}^{\infty}a_{2r}\sum_{k = r}^{\infty}\frac{(-1)^k(2\pi)^{2k} }{(2k - 2r + 1)!} + \sum_{r = 0}^{\infty}a_{2r + 1}\sum_{k = r + 1}^{\infty}\frac{(-1)^k(2\pi)^{2k} }{(2k - 2r)!}$$

$$ = \sum_{r = 0}^{\infty}a_{2r}\sum_{k = 0}^{\infty}\frac{(-1)^{k + r}(2\pi)^{2k +2r}}{(2k + 1)!} + \sum_{r = 0}^{\infty}a_{2r + 1}\sum_{k = 0}^{\infty}\frac{(-1)^{k + r + 1}(2\pi)^{2k + 2r + 2} }{(2k + 2)!}$$

$$ = \sum_{r = 0}^{\infty}a_{2r}(-1)^r(2\pi)^{2r - 1}\sum_{k = 0}^{\infty}\frac{(-1)^{k}(2\pi)^{2k + 1}}{(2k + 1)!} + \sum_{r = 0}^{\infty}a_{2r + 1}(-1)^r(2 \pi)^{2r}\sum_{k = 0}^{\infty}\frac{(-1)^{k + 1}(2\pi)^{2k + 2} }{(2k + 2)!}$$

$$ = \sum_{r = 0}^{\infty}a_{2r}(-1)^r(2\pi)^{2r - 1}\sin(2 \pi) + \sum_{r = 0}^{\infty}a_{2r + 1}(-1)^r(2 \pi)^{2r}(\cos(2 \pi) - 1)$$

$$ = \sum_{r = 0}^{\infty}a_{2r}(-1)^r(2\pi)^{2r - 1} \cdot 0 + \sum_{r = 0}^{\infty}a_{2r + 1}(-1)^r(2 \pi)^{2r} \cdot 0 = 0$$

It seems strange that the sum evaluates to $0$.

Reactant
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    Without having checked in detail whether your rearrangements are valid given your absolute convergence assumption, I'm not sure why you find it strange that the sum evaluates to $0$. One of the two sums is independent of $a_r$; you rearranged to collect all terms including $a_r$ and found that they add up to $0$. Why shouldn't they? – joriki Jul 09 '15 at 08:09
  • @jokiri Upper limit $2k$ was typo. – Reactant Jul 09 '15 at 08:19
  • @jokiri: It seems strange because I when selected specific $(a_n)$ sum should have been positive. – Reactant Jul 09 '15 at 08:24
  • I agree with joriki, the sums along the "$k$-axis" are null (since they correspond to DSE of sin/cos on values that nullified those functions), and since $a_r$ is constant along the "$k$-axis", I don't see a problem with that. – Nihl Jul 09 '15 at 08:25
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    I don't see why you should be able to choose $(a_n)$ so as to make the sum positive. It's got minus signs that vary with $k$, not with $r$, so each $a_r$ is multiplied by both positive and negative coefficients, and you can't change that by choosing $(a_n)$ with particular signs. It seems to me that if the sum does evaluate to $0$, your calculation does a perfect job of explaining why that is :-) – joriki Jul 09 '15 at 08:26

1 Answers1

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Another similar way to see that it works $$\sum_{r\geq0}\sum_{2k\geq r}\frac{\left(-4\pi^{2}\right)^{k}a_{r}}{\left(2k-r+1\right)!}=\sum_{r\geq0}a_{r}\left(-4\pi^{2}\right)^{-r/2}\sum_{m\geq0}\frac{\left(-4\pi^{2}\right)^{m/2}}{\left(m+1\right)!}=\frac{1}{2\pi i}\sum_{r\geq0}a_{r}\left(-4\pi^{2}\right)^{-r/2}\sum_{m\geq1}\frac{\left(2\pi i\right)^{m}}{m!} $$ $$=\frac{1}{2\pi i}\sum_{r\geq0}a_{r}\left(-4\pi^{2}\right)^{-r/2}\left(e^{2\pi i}-1\right)=0. $$

Marco Cantarini
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