Series and alternating series of the same function that converge to inverse quantities

Question

I know that from the definition for exponential we have that $e^x=\sum_{k\ge 0}\frac{x^k}{k!}$. As a consequence

$$e=\sum_{k\ge0}\frac{1}{k!}\quad\quad \text{and}\quad\quad \frac{1}{e}=\sum_{k\ge0}\frac{(-1)^k}{k!}$$

So Im curious about if exist some more functions that hold

$$f(k):\quad c=\sum_{k}f(k)\quad \text{and }\quad \frac{1}{c}=\sum_{k}(-1)^k f(k)\quad c\in\mathbb R\setminus \{1\}$$

P.S.: Im not sure about the tags for this question.

well... yes @Martigan, I forget to exclude the case for $c=1$, Thank you anyway. — Masacroso, Jun 19 '15 at 13:39
A possible idea: take any entire odd function f, write $g=e^f$, and look at the power series expansion of $g$ at 1 and -1. — Clement C., Jun 19 '15 at 13:54
Re my comment above: checking with some examples, this may not be alternating in every case... (only ensures series with the condition $g(-1)=1/g(1)$) — Clement C., Jun 19 '15 at 14:02

score 2 · Accepted Answer · answered Jun 19 '15 at 13:45

There are an infinity of cases as soon as you accept $f(k)$ to be whatever you want.

For instance$f(0)=a$, $f(1)=b$, $f(k>1)=0$.

Then every $(a,b)$ that fits $a^2=b^2+1$ are going to give $c=a+b$ and $\dfrac 1c=a-b$ as good answers.

For instance $a=2$ and $b=\sqrt 3$ will give $c=2+\sqrt 3$ and $\dfrac 1c=2-\sqrt 3$

1 Answers1