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?

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

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$
    – Valtteri
    Feb 23, 2013 at 3:03
  • $\begingroup$ Yes, that's right. $\endgroup$ Feb 23, 2013 at 3:06
  • $\begingroup$ Very good, I think I am convinced :D $\endgroup$
    – Valtteri
    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

$\ \ a_1=.1$

$\ \ a_2=.0101$

$\ \ a_3=.00001001$

$\ \ a_4=.0000000010001$

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


  • $\begingroup$ If I'm not wrong, you're adding $2^n$ zeros and $n$ zeroes? in the middle of the $1$s? $\endgroup$
    – Pedro
    Feb 23, 2013 at 2:53
  • 2
    $\begingroup$ @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). $\endgroup$ Feb 23, 2013 at 2:56
  • $\begingroup$ @PeterTamaroff It jumps from one $0$ to four $0$ from $a_2$ to $a_3$. $\endgroup$
    – awllower
    Feb 23, 2013 at 2:57
  • $\begingroup$ OK, but after that you add $n-1$ zeroes and a $1$. $\endgroup$
    – Pedro
    Feb 23, 2013 at 2:58
  • $\begingroup$ @awllower Yes, I overlooked that. $\endgroup$
    – Pedro
    Feb 23, 2013 at 2:58

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