# Does every sequence of rationals, whose sum is irrational, have a subsequence whose sum is rational

Assume we have a sequence of rational numbers $a=(a_n)$. Assume we have a summation function $S: \mathscr {L}^1 \mapsto \mathbb R, \ \ S(a)=\sum a_n$ ($\mathscr {L}^1$ is the sequence space whose sums of absolute values converges). Assume also that $S(a) \in \mathbb R \setminus \mathbb Q$.

I would like to know if every such sequence $a$ has a subsequence $b$ (infinitely long) such that $S(b) \in \mathbb Q$.

Take as an example $a_n = 1/n^2$. Then $S(a)=\pi^2/6$. But $a$ has a subsequence $b=(b_n)=(1/(2^n)^2)$ (ie. all squares of powers of $2$). Then $S(b)=4/3$. Is this case with every such sequence?

• I think $S(b)$ should be $4/3$? Feb 23, 2013 at 2:49
• Interesting question!
– Pedro
Feb 23, 2013 at 2:50

No. For example, take the sequence $a_n=2^{-2^n}$, $n=1$, $2$, $\dots$. An infinite subsequence $(a_{n_k})$ of $(a_n)$ will have sum $$S:=\sum_k a_{n_k}=\sum_k 2^{-2^{n_k}}.$$ So, the binary expansion of $S$ will have $1$s in positions $2^{n_1}$, $2^{n_2}$, $2^{n_3}$, $\dots$, and $0$s everywhere else. This is not a periodic sequence, so $S$ must be irrational.

• Looks cool. So you convert it to binary and see that the decimal expansion doesn't repeat ever, so it is irrational and same if you take any subsequence? Feb 23, 2013 at 3:03
• Yes, that's right. Feb 23, 2013 at 3:06
• Very good, I think I am convinced :D Feb 23, 2013 at 3:07

No; for example, if $(n_i)$ is a strictly increasing sequence of positive integers, then we can imitate the proof of the irrationality of $e$ to see that

$$\sum_{i=1}^\infty \frac{1}{n_1 \dots n_i} \notin \mathbf Q.$$

But every sub-series of this series has the same property (it just amounts to grouping some of the $n_i$ together).

What if you took

$\ \ a_1=.1$

$\ \ a_2=.0101$

$\ \ a_3=.00001001$

$\ \ a_4=.0000000010001$

$\ \ \ \ \ \ \ \ \vdots$

?

• If I'm not wrong, you're adding $2^n$ zeros and $n$ zeroes? in the middle of the $1$s?
– Pedro
Feb 23, 2013 at 2:53
• @PeterTamaroff The first nonzero digit of $a_{n+1}$ occurs after the last nonzero digit of $a_n$. The "length of the zeroes in the middle" increase (so as to force any sum of a subsequence to not have a repeating decimal expansion). Feb 23, 2013 at 2:56
• @PeterTamaroff It jumps from one $0$ to four $0$ from $a_2$ to $a_3$. Feb 23, 2013 at 2:57
• OK, but after that you add $n-1$ zeroes and a $1$.
– Pedro
Feb 23, 2013 at 2:58
• @awllower Yes, I overlooked that.
– Pedro
Feb 23, 2013 at 2:58