Show that the numbers of the form $\sum_{k=1}^{\infty} \frac{a_j}{3^k}$ , where $a_j = 0$ or $a_j = 1$ is countable .
If $A = \cap_i^n A_1$ is countably infinite, then atleat one $A_i$ is counntable.
I know that the series $\sum_{k=1}^{\infty} \frac{a_j}{3^k}$ is convergent , because $\sum_{k=1}^{\infty} \frac{a_j}{3^k} \leq \sum_{k=1}^{\infty} \frac{1}{3^k}$, I want to prove that every series is convergent to rational number, but i am unable.
For second statement , I think it is false, please give me any counter example
any help would be appreciated. Thank you
Both statements are false, actually.
For (1): You've already shown that every such series converges; it's not hard to show that any two series which differ at some point (e.g. $a_n\not=b_n$ for some $n$) converge to different reals. Thus, you're really just counting the number of sequences of possible $a_i$s.
Using this, can you construct a bijection between the set of reals which are representable by such a series, and some set you already know is uncountable? (Or, apply a diagonal argument directly.)
For (2): Can you think of two uncountable sets whose intersection is "very small"?