Conjecture: There do not exist functions $f, g : \mathbb N \to \mathbb R^+$ and $c \in \mathbb R\setminus\{0\}$ such that simultaneously

  • $S_0=\sum_{n=1}^{\infty}\frac{(-1 )^n}{f(n)} = 0$
  • $S_1=\sum_{n=1}^{\infty}\frac{(-1 )^n}{f(n)}*g(n)^c = 0$
  • $S_2=\sum_{n=1}^{\infty}\frac{(-1 )^n}{f(n)}*\frac{1}{g(n)^c} = 0$

It seems correct but I could not find a proof or counterexample to this. Can anyone please help?

As pointed out if $g(n)=1$ then its trivial. Thus, we are concerned where $|g(n)^c| \neq |\frac{1}{g(n)^c}|$.

  • $\begingroup$ What about $g(n)=1$? $\endgroup$ – Crostul Feb 28 '19 at 7:27
  • $\begingroup$ I was going to add that one trivial case not allowed. :) updating. $\endgroup$ – TheoryQuest1 Feb 28 '19 at 7:28
  • $\begingroup$ $n\in\mathbb{N}$, no? $\endgroup$ – uniquesolution Feb 28 '19 at 7:34
  • $\begingroup$ Similarly, what about $g(n)$ is a constant itself? $\endgroup$ – Euler....IS_ALIVE Feb 28 '19 at 7:36
  • $\begingroup$ yes. typo error. $\endgroup$ – TheoryQuest1 Feb 28 '19 at 7:36

Let $f(n)=n^n$ for odd $n$ and $(n-1)^{(n-1)}$ for even $n$. Every second term in $f(n)$ is the same and hence the summation converges to $0$. Now let $g(n)=2$.

Obviously both summation converge to $0$ as they are a constant multiple of the summation of all $\frac{1}{f(n)}$. This disproves the original statement.

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  • $\begingroup$ do they both converge to 0? $\endgroup$ – TheoryQuest1 Feb 28 '19 at 7:46
  • $\begingroup$ Sorry, I've corrected it. $\endgroup$ – Peter Foreman Feb 28 '19 at 8:01
  • $\begingroup$ The original poster stated that $g(n)$ is a constant cannot be allowed because of triviality. $\endgroup$ – Euler....IS_ALIVE Feb 28 '19 at 8:03
  • $\begingroup$ The post states "we are concerned where $|g^c(n)|\ne |\frac{1}{g^c(n)}|$" which for this chosen function is true. $\endgroup$ – Peter Foreman Feb 28 '19 at 8:08
  • $\begingroup$ Thank you but If we were to assume $g(n)$ as some constant $k$, then it amounts to multiplying the series $S_0$ or 0 with $k^c$ and $1/k^c$ respectively. And that is obviously 0. Thus, we are ignoring all such trivial cases. $\endgroup$ – TheoryQuest1 Feb 28 '19 at 8:26

Seems to me that your conjecture is false.

Let $h(x)$ be a function given by the power series $$h(x)=\sum_{n=1}^{\infty}a_nx^n$$ Assume that $h(x)$ is well defined, and that $h(1/2)=h(1)=h(2)=0$. (For example take $h(x)= (x-1/2)(x-1)(x-2)k(x)$ for a suitable $k(x)$). For those $n$ such that $a_n\neq 0$, put $f(n)=(-1)^n/a_n$, and take $g(n)=2^{-n}$. Then $$\sum_{n=1}^{\infty}\frac{(-1)^n}{f(n)}g(n)=h(1/2)=0$$ and $$\sum_{n=1}^{\infty}\frac{(-1)^n}{f(n)}\frac{1}{g(n)}=h(2)=0$$ and also $$\sum_{n=1}^{\infty}\frac{(-1)^n}{f(n)}=h(1)=0$$

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