what I ask is a bit convoluted. First, we will define a relace:
Def: Let $\;{q_1},{q_2}\dots,{q_n}\dots \; ({q_i}\in\Bbb Q)$ such that $\beta=\sum_{i=1}^\infty q_i$. We say that a relace of the series is a new series $\sum_{j=1}^\infty p_j$ fulfilling:
- $p_j=\sum_{i\in I_j} q_i$,$\,$ with $\,$ $\Bbb N =\bigsqcup_{j\in\Bbb N} I_{j}\;$ (disjoint union)
- $\beta=\sum_{j=1}^\infty p_j$
If the relace $\sum_{j=1}^\infty p_j$ is a geometric series, we say that it's a geometric relace.
Example: Let $M_e$ be the Mengoli series: $$M_e=\sum_{n=1}^\infty \frac{1}{n(n+1)}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\dots$$ and we choose $I_k=\{2^{k-1},\dots,2^k-1\}$,$\;$ for $k=1,2,3,\dots$ Then $$M_e=(\frac{1}{2})+(\frac{1}{6}+\frac{1}{12})+(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56})+(\frac{1}{72}+\dots=(\frac{1}{2})+(\frac{1}{4})+(\frac{1}{8})+\dots$$ which is a geometric series: $$M_e=\sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}=2^{-1}\frac{1}{1-2^{-1}}=1$$
I want to prove this theorem:
Theorem: Let $\beta=\sum_{i=1}^\infty q_i$ be a rational series (of rational terms). Then the following statements are equivalent:
- $\beta$ admits a geometric relace.
- $\beta\in\Bbb Q$
Proof:
- $\Rightarrow$ 2) is evident, since every geometric series (of rational terms) converges to a rational number.
- $\Rightarrow$ 1) ? (I don't even know where to start)
Thanks for trying it!
We are going to consider the series $$ {1\over2}+{1\over3}+{1\over7}+{1\over43}+\dotsb=1 $$ where each denominator is one more than the product of all the previous denominators, e.g., $43=1+(2\times3\times7)$. Note also that $$ 1={1\over2}+{1\over2}={1\over2}+{1\over3}+{1\over6}={1\over2}+{1\over3}+{1\over7}+{1\over42}=\dotsb $$ so each successive denominator is larger by $1$ than the denominator that would make the terms sum to $1$.
Now we are aiming for a geometric series with first term $a$ and common ratio $r$, and sum $1$, so $$ a+ar+ar^2+\dotsb={a\over1-r}=1 $$ From this it follows that $r=1-a$.
Now in any geometric relace the first term will be the largest, so it must include the term $1/2$, since the series can't have two terms greater than or equal to one-half. So, $a\ge1/2$.
Then the second term, $ar=a(1-a)$, will be at most $1/4$, so neither it, nor any succeeding term, can include $1/3$. So $1/3$ must be in the first term, so $a\ge(1/2)+(1/3)=5/6$.
From $a\ge5/6$, we get for the second term $a(1-a)\le (5/6)(1/6)=5/36<1/7$, so $1/7$ can't be in the second (or any later) term, so it too must be in the first term. Thus, $a\ge(1/2)+(1/3)+(1/7)=41/42$.
Then $a\ge41/42$ implies the second term $a(1-a)\le(41/42)(1/42)<1/43$ (because $41\times43<42^2$), so $1/43$ must be in the first term. At this point, we should make this a proof by mathematical induction, but I think the argument is clear enough from the preceding calculations; every term in our series must be included in the first, and thus only, term of the alleged geometric relace, so there is no geometric relace.