Prove that $ \sum \frac{2^k}{k}$ is divisible by $2^M$ For each integer $M > 0$, there ${\bf exists}$ an $n$ such that
$$ \sum_{k=1}^n \dfrac{ 2^k}{k} $$
is divisible by $2^M$
${\bf try}$ Im struggling a bit to visualize this exercise. So, I tried to see for concrete number, for instance take $M=1$, then $n=2$ works: as
$$ 2 + \dfrac{2^1}{2} = 2^1 (1 + 2 )$$
Now, take $M=2$ and factor
$$ 2^{2} \underbrace{ \left( \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{2^4}{4} ... + \dfrac{2^{n-2} }{n} \right) }_{(*)}$$
now, we need to choose $n$ so that $(*)$ is an integer. Im unable to do so. Any help?

Right so the "elementary answer." There are actually a couple that I've seen. One is cited by Gouvêa in his p-adic numbers book (Gouvêa, like me, also said the problem was too difficult to work out the answer from scratch).
The solution Gouvêa cites is Exercise 10.10 of D. P. Parent Exercises in Number Theory (1984).
The rough idea there (assuming I'm summarizing correctly) is
$$ 0 = \frac{1 - (1 - 2)^{2^n}}{2^n} = \sum_{k = 1}^{2^n} (-1)^{k+1}2^k \frac{1}{2^n}\binom{2^n}{k} \equiv \sum_{k=1}^{2^n} \frac{2^k}{k} \pmod{2^h} $$
for $n \ge h$. Plus there's some other work to compare this with the partial sums that are not powers of $2$.

The next two solutions I found cited by the OEIS (https://oeis.org/A087910). Namely the two solutions given for the 2002 Sydney University Mathematical Society Problems Competitions Problem 9.
Solution 1 Summary
If $n$ is even then
$$ 1 = (-1)^n = (1 - 2^n) = \sum_{k = 0}^n \binom{n}{k}(-2)^k. \tag{1} $$
Subtract 1 and divide by $n$ to get
$$ 0 = \sum_{k = 1}^n \frac1n \binom{n}{k}(-2)^k. $$
Then
$$ \frac1n \binom{n}{k} = \frac{(n-1)(n-2)\cdots(n-k+1)}{k!} = \frac{(-1)^{k-1}}{k} + n\frac{m_{n,k}}{k!} \tag{2}$$
for some integer $m_{n,k}$ (separate the term $(-1)(-2)\cdots(-k+1)$ from all the terms divisible by $n$).
By $(1)$ and $(2)$,
$$ \sum_{k = 1}^n \frac{2^k}{k} = n \sum_{k = 1}^n \frac{(-2)^km_{n,k}}{k!}$$
Then you use a well known fact that $v_2(k)! \le O(k)$. Thus
$$ v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) = v_2(n) + v_2\left( \sum_{k = 1}^n \frac{(-2)^km_{n,k}}{k!} \right) \ge v_2(n). $$
Now we can see that when $n = 2^k$ this tends to infinity.

Solution 2 Summary
First show that
$$ \sum_{k = 1}^n \frac{2^k}{k} = \frac{2^n}{n} \sum_{k = 0}^{n-1} \frac{1}{\binom{n-1}{k}}. $$
Next, use the well-known formula $$v_2(n!) = \sum_{i = 0}^\infty \left\lfloor \frac n{2^i} \right\rfloor$$ to get
$$ v_2\left( \binom{n}{k} \right) = v_2(n!) - v_2(k!) - v_2((n - k)!) = \sum_{i = 0}^\infty \left\lfloor \frac n{2^i} \right\rfloor - \left\lfloor \frac k{2^i} \right\rfloor - \left\lfloor \frac {n-k}{2^i} \right\rfloor. $$
Then by some analysis,
$$v_2\left( \binom{n-1}{k} \right) \le r \text{ if } 2^r + 1 \le n \le 2^{r + 1}. $$
So if $n$ is even and $2^r + 1 \le n \le 2^{r + 1}$ then
$$ v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) \ge n - r. $$
And the result follows.

You'll have to see the solutions I linked to if you want all the details, they wouldn't fit in one answer. I hope this gives you some appreciation for the $2$-adic logarithm approach.
