Prove that $ \sum \frac{2^k}{k}$ is divisible by $2^M$ For each integer $M &gt; 0$, 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?
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 &quot;more elementary&quot; 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.
This certainly doesn't seem true
If any $n &gt; 2$ there is a prime $p$ so that $2&lt;p \le n &lt; 2p$ (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 $n &gt; 2$.
..... unless I'm making a boneheaded error.