A true statement, which should not be too complicated to prove is the following:
For each $M >0$ the exists an $n$ such that the partial sum: $$q_n = 2+\frac{2^2}{2}+\frac{2^3}{3}+\dots+\frac{2^n}{n}$$ is divisible by $2^M$.
If we interpret $q_n$ in the p-adic numbers $\mathbb{Q}_2$, a possible proof can be achieved, using the MacLaurin series for the logarithm of $1-x$: $$\log (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots $$ Since powers of 2 are ''small'' in $\mathbb{Q}_2$, it turns out that we can plug $x=2$ to compute the logarithm of $-1$: $$\log (-1) = \log(1-2) = -\left(2 + \frac{2^2}{2} + \frac{2^3}{3} + \cdots \right)$$ Now as the series converges in $\mathbb{Q}_2$, it must converge to zero, by the usual properties of the logarithm: $$2\log(-1) = \log(1) = 0$$ This means that the partial sums in $\mathbb{Q}_2$ $$2 + \frac{2^2}{2} + \frac{2^3}{3} + \dots +\frac{2^n}{n}$$ must get closer and closer to zero as $n$ grows. Then the p-adic expansion begin with longer and longer stretches of zeros, the result follows. But the question is how to prove this muddled arithmetical fact without resorting to p-adic numbers, can an equally simple proof be constructed? How can $n(M)$ be estimated?
