Let $a_n$ denote the exponent of $2$ in the numerator of $\sum_{i=1}^n \dfrac{2^i}{i}$ when written in the lowest form. For example, $a_1=1,a_2=2,a_3=2.$ Prove that $\lim\limits_{n\to\infty} a_n=\infty$.
Source: problem 11 from this problem set.
One could try to compute the first few values of the expression, but it's very unlikely that there's a pattern in the terms of the sequence. We obviously just need to find an unbounded subsequence (e.g. a strictly increasing subsequence). There probably isn't an explicit formula for $\sum_{i=1}^n \dfrac{2^i}i$. The denominator of the fraction would clearly divide $n!,$ and by Legendre's formula, $v_p(n!) = \sum_{i=1}^\infty \lfloor n/p^i\rfloor$ for any prime number p and positive integer n. We clearly have that $\sum_{i=1}^n \dfrac{2^i}i = \dfrac{\sum_{i=1}^n 2^i\cdot n!/i}{n!}$. Also, there are several basic properties of $v_p(n)$ that could be useful.Let $K_n = \sum_{i=1}^n \dfrac{2^i}i$ for all n. Then $K_1 = 2, K_2 = 1/2(2\cdot 2 + 4\cdot 1) = 4.$ The hint to the problem says that the series equals $-\log(1-x)$ for $x=2$ in $\mathbb{Z}_2$, though how does this even make sense when $-\log(1-x)$ isn't defined for $x<0$?
