How can I find a sequence of numbers $A_n \in \mathbb{R} $ (not infinitely many of them being zero) such that

$ \sum_{n=1}^{\infty} (-1)^{n-1} \cdot A_n = 1$


$ \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \sqrt{\frac{1-n \cdot (A_n)^{2}}{n}} = 0 \space \space \space $ ?

( It is also required that the latter series doesn’t have infinitely many zeroes )

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    $\begingroup$ Why do you want to know this? Is there some background to this? $\endgroup$ – Kenta S Jul 7 '18 at 0:59
  • $\begingroup$ Those sequences appeared to me while I was studying the Dirichlet Eta function (alternate Zeta). $\endgroup$ – Leonardo Bohac Jul 7 '18 at 1:07
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    $\begingroup$ Some of them could, but not all. I understand your second non-conviction, and I agree that the second series could be of complex numbers, though I’m assuming that there is a real valued series also, and that is the one I’m seeking. I have my work written in my papers, I’ll get to Latex them. $\endgroup$ – Leonardo Bohac Jul 7 '18 at 1:42
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    $\begingroup$ Have you tried taking $A_n=1/\sqrt n$ for $n \geq 3$ and solving the two equations to find $A_1$ and $A_2$? $\endgroup$ – Kavi Rama Murthy Jul 7 '18 at 2:01
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    $\begingroup$ You can pick $A_n=$*anything* (for example $A_n=2^{-n}$) for $n \ge 3$, and then solve for $A_1, A_2$. Obviously, these two numbers won't have a nice closed form, but they exist. $\endgroup$ – Crostul Jul 7 '18 at 17:43

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