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?

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.
