I've read about normal numbers and wonder if each normal number contains each sequence of digits infinitely often. I think the answer is rather straightforward, isn't it, and maybe too obvious to be stated explicitely:
If a normal number contains the sequence $d_1d_2\dots d_k$ it also contains every sequence $d_1d_2\dots d_k\dots d_n$ of which there are infinitely many, thus it contains $d_1d_2\dots d_k$ infinitely often.
The definition of "normal" (in base 10) is that the decimal expansion of the number must contain copies of the sequence $d_1\ldots d_k$ with limiting density $10^{-k}$. If there were only finitely many copies, the limiting density would be $0$, so the number would not be normal.