# Conjecture regarding alternating infinite series

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 $$|g(n)^c| \neq |\frac{1}{g(n)^c}|$$.

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

• do they both converge to 0? – TheoryQuest1 Feb 28 '19 at 7:46
• Sorry, I've corrected it. – Peter Foreman Feb 28 '19 at 8:01
• The original poster stated that $g(n)$ is a constant cannot be allowed because of triviality. – Euler....IS_ALIVE Feb 28 '19 at 8:03
• The post states "we are concerned where $|g^c(n)|\ne |\frac{1}{g^c(n)}|$" which for this chosen function is true. – Peter Foreman Feb 28 '19 at 8:08
• 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. – 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 $$\sum_{n=1}^{\infty}\frac{(-1)^n}{f(n)}=h(1)=0$$