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?
 A: First, an important definition

Definition the $2$-adic valuation $v_2$ is given by
$$v_2\left( 2^n\frac{p}{q} \right) = n$$
where $p, q$ are odd (not divisible by $2$).

For instance $v_2(12/5) = 2$ since we can write $12/5$ as $2^2 \frac{3}{5}$.
In this language, what we want to show is that
$$\lim_{n \to \infty} v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) = \infty. \tag{$*$}$$

In my opinion the best/most conceptual/requires the least tricks way to prove this is to identify the infinite series
$$ \sum_{k = 1}^\infty \frac{2^k}{k} $$
as the $2$-adic logarithm of $-1$. That is, it is $-\log(1 - 2) = -\log(-1)$. Recall that the logarithm is given by the series
$$-\log(1 - x) = \sum_{k = 1}^\infty \frac{x^k}{k}. $$
Given this, and assuming that basic properties of logarithms still hold in the $2$-adics, we have
$$\log(-1) + \log(-1) = 2 \log(-1) = \log((-1)^2) = \log(1) = 0. $$
Therefore $\log(-1) = 0$ as a $2$-adic number.
Well the series converges to $0$ in the $2$-adics if and only if the partial sums grow to $\infty$ in $2$-adic valuation. I.e. The series converges to $0$ if and only if $(*)$ holds.
This requires some work to understand:

*

*what does it mean to converge in the $2$-adic valuation

*do these series converge?

*is $\log(xy) = \log x + \log y$ still true and when?

For all these questions, I must refer to Keith Conrad's notes. The identity $(*)$ is discussed at the top of page 29 (Example 8.10) and I'm not going to fit 29 pages of exposition in here.
However, I will say that this approach is much nicer than the non-logarithm approach. Because that approach requires several very hard-to-find tricks. If I were locked in a room with no internet or any other references and I had to prove this fact, I would pick the p-adic logarithm approach because at least then I have some idea what needs to be done.

I'll try to find a "more elementary" approach in a separate answer (this one's getting sort of long). But I'll warn you now: don't expect it to be easy to understand or enlightening.
A: 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.
A: This certainly doesn't seem true
If any $n > 2$ there is a prime $p$ so that $2<p \le n < 2p$ (Bertands postulate) and so
And so if $2^m| \sum_{k=1}^n \frac {2^k}k$ then $\sum_{k=1}^n \frac {2^k}k=W$ is as an integer.  And so $W-\frac {2^p}p$ is not an integer.
But $W - \frac {2^p}p= \frac {pW - 2^p}p$ is a fraction in lowest terms.
Which means $\sum_{k=1;k\ne p}^n \frac {2^k}k = \frac {pW -2^p}p$.  But be figuring out the least common denominator of $\frac {2^k}{k}; k\ne p$ we get the lets common denominater is $\operatorname{lcm}(k; k\ne p)$.  But that will not reduce so $\frac{pW-2^p}p$ unless $p$ divides into $\operatorname{lcm}(k; k\ne p)$.
Which it clearly can't as $\operatorname{lcm}(k; k\ne p)$ does not have $p$ as a prime divisor of any of the $k$ terms.
SO.... unless I'm making a bonehead error, this is not possible.
Not only is this not true  $\sum_{k=1}^n \frac {2^k}k$ is never an integer if $n > 2$.
..... unless I'm making a boneheaded error.
