Problem statement here.
$a_n$ is the number of odd digits in $2^n$ (in base $10$). $b_n$ is the number of digits in $2^n$.
For part A)
We have the following curious reciprocity property.
The parity of the $j^{th}$ digit (from the right, starting with $j=0$) of $2^n$ is the same as the $n^{th}$ digit after the decimal place of $10^{-j}$ in base-2.
For eg, $10^{-2} = .00000010100011110101110000\dots$ and the first few powers of $2$ are
$2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$
and the corresponding parities of the seconds digits (note starting with $0$).
$0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1$ which exactly matches the first few digits of $10^{-2}$.
The proof is pretty simple. We have that the parity of the $j^{th}$ digit of $2^n$ is basically
$$ \lfloor 2^n 10^{-j} \rfloor \mod 2$$.
Which is exactly the same as the $n^{th}$ digit of $10^{-j}$ in base-2. $2^n$ just shifts the decimal point of $10^{-j}$ right by $n$ places and that digit gives us the parity of $ \lfloor 10^{-j} 2^n \rfloor$.
Call this number $f(n, j)$. Thus
$$\sum_{n=1}^{\infty} \frac{a_n}{2^n} = \sum_{n=1}^{\infty} \sum_{j=0}^{\infty} \frac{f(n,j)}{2^n}$$
Exchanging the order of summation gives
$$ \sum_{j=0}^{\infty} \sum_{n=1}^{\infty} \frac{f(n,j)}{2^n}$$
By the above reciprocity property, we have that $\sum_{n=1}^{\infty} \frac{f(n,j)}{2^n} = 10^{-j}$.
Thus the sum is essentially $\sum_{j=0}^{\infty} 10^{-j} = \frac{1}{9}$.
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For Part B)
To show that $S = \sum \frac{b_n}{2^n}$ is irrational.
Consider the positions where the number of digits in $2^n$ increases. Define $d_n = 1$ if
$b_n \gt b_{n-1}$ and $d_n = 0$ otherwise.
By considering $2S$ and subtracting we see that it is enough to show that
$$ \sum \frac{d_n}{2^n}$$ is irrational.
$d_n = 1$ iff $n-1$ is the integer part of $ k \log_2(10)$ for some $k \in N$, because then we have that $2^n \lt 10^k \lt 2^{n+1}$.
Since $\log_2(10)$ is irrational the sequence of $0$'s and $1$'s which is $d_n$ is infinite and non-periodic* and hence the number $\sum \frac{d_n}{2^n}$ is irrational when we look at it as a number in base-2.
*It is non-periodic because the sequence $\lfloor k \gamma\rfloor, k \in N$ for some irrational $\gamma$ has an irrational density. Which will not be the case if it were periodic.
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For Part C).
If we look at the $n$ such that $d_{n+1} = 1$, then we see that $n$ is a multiple of 3 for a good few terms (till 99) and then becomes $101$.
The number in question is what we get if we assume that the pattern of multiples of 3 continues, so 102 instead of 101. Since we get 102 instead of 101, the irrational number is larger.
$a_n$ is the number of odd digits in $2^n$ (in base $10$). $b_n$ is the number of digits in $2^n$.
For part A)
We have the following curious reciprocity property.
The parity of the $j^{th}$ digit (from the right, starting with $j=0$) of $2^n$ is the same as the $n^{th}$ digit after the decimal place of $10^{-j}$ in base-2.
For eg, $10^{-2} = .00000010100011110101110000\dots$ and the first few powers of $2$ are
$2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$
and the corresponding parities of the seconds digits (note starting with $0$).
$0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1$ which exactly matches the first few digits of $10^{-2}$.
The proof is pretty simple. We have that the parity of the $j^{th}$ digit of $2^n$ is basically
$$ \lfloor 2^n 10^{-j} \rfloor \mod 2$$.
Which is exactly the same as the $n^{th}$ digit of $10^{-j}$ in base-2. $2^n$ just shifts the decimal point of $10^{-j}$ right by $n$ places and that digit gives us the parity of $ \lfloor 10^{-j} 2^n \rfloor$.
Call this number $f(n, j)$. Thus
$$\sum_{n=1}^{\infty} \frac{a_n}{2^n} = \sum_{n=1}^{\infty} \sum_{j=0}^{\infty} \frac{f(n,j)}{2^n}$$
Exchanging the order of summation gives
$$ \sum_{j=0}^{\infty} \sum_{n=1}^{\infty} \frac{f(n,j)}{2^n}$$
By the above reciprocity property, we have that $\sum_{n=1}^{\infty} \frac{f(n,j)}{2^n} = 10^{-j}$.
Thus the sum is essentially $\sum_{j=0}^{\infty} 10^{-j} = \frac{1}{9}$.
-------------
For Part B)
To show that $S = \sum \frac{b_n}{2^n}$ is irrational.
Consider the positions where the number of digits in $2^n$ increases. Define $d_n = 1$ if
$b_n \gt b_{n-1}$ and $d_n = 0$ otherwise.
By considering $2S$ and subtracting we see that it is enough to show that
$$ \sum \frac{d_n}{2^n}$$ is irrational.
$d_n = 1$ iff $n-1$ is the integer part of $ k \log_2(10)$ for some $k \in N$, because then we have that $2^n \lt 10^k \lt 2^{n+1}$.
Since $\log_2(10)$ is irrational the sequence of $0$'s and $1$'s which is $d_n$ is infinite and non-periodic* and hence the number $\sum \frac{d_n}{2^n}$ is irrational when we look at it as a number in base-2.
*It is non-periodic because the sequence $\lfloor k \gamma\rfloor, k \in N$ for some irrational $\gamma$ has an irrational density. Which will not be the case if it were periodic.
------------
For Part C).
If we look at the $n$ such that $d_{n+1} = 1$, then we see that $n$ is a multiple of 3 for a good few terms (till 99) and then becomes $101$.
The number in question is what we get if we assume that the pattern of multiples of 3 continues, so 102 instead of 101. Since we get 102 instead of 101, the irrational number is larger.
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