$\bf{Problem (a) }$
Let $p_n $ denote the nth prime.
Show that the series $\sum_{n=1}^{\infty} \frac{p_n}{10^{2^n}} $ converges.
we know that $ p_n \leq 4^n $ by chebychev additionally $4^n \leq 4^{2^{n}} $ so we have that $\sum_{n=1}^{\infty} \frac{p_n}{10^{2^n}}\leq \sum_{n=1}^{\infty} (\frac{4}{10})^{2^{n}} $ by ratio test we have that $\frac{(\frac{4}{10})^{2^{n+1}}}{(\frac{4}{10})^{2^{n}} } = (\frac{4}{10})^2 <1$ so the series converges by ratio test.
$\bf{Problem (b) }$
let $\alpha =\sum_{n=1}^{\infty} \frac{p_n}{10^{2^n}} $ show that $p_n = \lfloor 10^{2^n} \alpha \rfloor - 10^{2^{n-1}} \lfloor 10^{2^{n-1}} \alpha \rfloor$
It gives the following hint; Prove $0 < 10^{2^n}\sum_{m=n+1}^{\infty} \frac{p_m}{10^{2^m}}<1$ but im not really sure how this applies but i think its a fascinating result as it givens the nth prime i am assuming we have no idea what $\alpha$ is?