### Instruction: Compute the smallest positive integer $n$ such that
\[\sum_{k = 0}^n \log_2 \left( 1 + \frac{1}{2^{2^k}} \right) \ge 1 + \log_2 \frac{2014}{2015}.\].### Response: The problem asks me to find the smallest $n$ that makes a certain sum of logarithms greater than or equal to a given expression.
I notice that the base of the logarithms is 2, which suggests that binary representations or powers of 2 might be involved.
I also notice that the argument of each logarithm in the sum is of the form $1 + \frac{1}{2^{2^k}}$, which is a fraction with a power of 2 in the denominator and 1 more than that in the numerator.
This makes me think of the binary representation of numbers between 0 and 1, where each digit after the decimal point corresponds to a power of 2 in the denominator.
For example, $\frac{1}{2} = 0.1_2$, $\frac{1}{4} = 0.01_2$, $\frac{1}{8} = 0.001_2$, and so on.
So, if I write $1 + \frac{1}{2^{2^k}}$ in binary, I get $1.0\underbrace{00\cdots 0}_{2^k \text{ zeros}}1_2$.
This means that $\log_2 \left( 1 + \frac{1}{2^{2^k}} \right)$ is the position of the second 1 in the binary representation of $1 + \frac{1}{2^{2^k}}$, which is $-2^k$.
Therefore, the sum $\sum_{k = 0}^n \log_2 \left( 1 + \frac{1}{2^{2^k}} \right)$ is just the sum of the negative powers of 2 from $-1$ to $-2^n$, which is the binary representation of $\frac{2^{n+1} - 1}{2^{n+1}}$.
So, the problem is equivalent to finding the smallest $n$ such that $\frac{2^{n+1} - 1}{2^{n+1}} \ge 1 + \log_2 \frac{2014}{2015}$.
I can simplify this inequality by multiplying both sides by $2^{n+1}$ and subtracting 1, which gives me $2^{n+1} - 2 \ge 2^{n+1} \log_2 \frac{2014}{2015}$.
Then, I can divide both sides by $2^{n+1}$ and take the logarithm base 2, which gives me $\log_2 \left( 1 - \frac{1}{2^{n+1}} \right) \ge \log_2 \frac{2014}{2015}$.
I can use the logarithm property that $\log_b a - \log_b c = \log_b \frac{a}{c}$ to rewrite this as $\log_2 \frac{2^{n+1} - 1}{2^{n+1}} \ge \log_2 \frac{2014}{2015}$.
Now, I can compare the fractions inside the logarithms and see that they are both less than 1, so the logarithm is negative and the inequality is reversed when I take the exponent of both sides.
This gives me $\frac{2^{n+1} - 1}{2^{n+1}} \le \frac{2014}{2015}$.
I can cross-multiply and simplify this to $2^{n+1} \le 2015$.
To find the smallest $n$ that satisfies this inequality, I can try some values of $n$ and see when the power of 2 exceeds 2015.
For example, $2^8 = 256$, $2^9 = 512$, $2^{10} = 1024$, $2^{11} = 2048$.
So, $n = 10$ is too small, but $n = 11$ is too big.
Therefore, the smallest $n$ that works is $n = 10$.