Assume we have a sequence of positive rational numbers $(a_n)$, and $\sum_{n=1}^\infty a_n = x$ and $x$ is transcendental. If we have a subsequence of $(a_n)$, $(b_n)$ and $\sum_{n=1}^\infty b_n = y$ and $y$ is not a rational number, is $y$ also transcendental? Further, if $y$ is transcendental, is it of the form $x-s,$ where $s$ is a rational number?
Take as an example the Basel problem. We have that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$ This is a transcendental number. The sum of every finite subsequence of $(\frac{1}{n^2})$ is rational and there are infinitely many infinite subsequences that sum to a rational number, for example, $(\frac{1}{(p^n)^{2}})$ where $p$ is a prime.
What if we have a subsequence whose sum is not a rational number? Say $(k_n)$ is such a sequence. Is the sum $\sum_{n=1}^\infty k_n$ transcendental? Is $\sum_{n=1}^\infty k_n = \frac{\pi^2}{6}-s$, where $s\in\mathbb Q$? Or could we select the members very carefully and get, say, $\sqrt{2}?$
I figured that if we see the reciprocals of squares as a set $A$, and let $B = \{z \in P(A)\ |\ \sum_{i \in z} i \in \mathbb Q\}$, then $B$ is an ideal, so the set of subsets whose sum is irrational ($P(a)\setminus B$) should be filter. But I doubt it is much use.
Best Answer
In this answer, I prove that there is a subsequence of $\sum \frac{1}{n^2}$ which sums to $x$, for any $x \in \left(0, \frac{\pi^2}{6} - 1\right] \cup \left(1, \frac{\pi^2}{6}\right]$. You can produce such a subsequence greedily: that is, at each step, always take the largest remaining term which doesn't cause you to exceed your target.
For example, this method gives us a series $$ \sqrt{2}=1+1/2^2+1/3^2+1/5^2+1/9^2 + 1/37^2+1/195^2+1/8584^2+1/1281816^2+\dots $$ assuming I haven't run into any roundoff errors.