How prove this $x_{2^n}\ge 2^n-n+1$

Question

the sequence $ (x_n)_{n\ge 1}$, $ x_n$ being the exponent of 2 in the decomposition of the numerator of $$\dfrac{2}{1}+\dfrac{2^2}{2}+\cdots+\dfrac{2^n}{n}$$ goes to infinity as $ n\to\infty$. Even more, prove that $$x_{2^n}\ge 2^n-n+1$$

My idea: maybe $$\dfrac{2}{1}+\dfrac{2^2}{2}+\cdots+\dfrac{2^n}{n}=\dfrac{2^n}{n}\cdot M'?$$ where $M'$ is postive numbers.

so we after we replacing $n$ by $2^n$, then $$\dfrac{2}{1}+\cdots+\dfrac{2^{2^n}}{2^n}=\dfrac{2^{2^n}}{2^n}M''=2^{2^n-n}M''$$ then I can't,Thank you for your help

Answer 1

Now,I have solution This nice problem, we only note that $$\dfrac{2}{1}+\dfrac{2^2}{2}+\cdots+\dfrac{2^n}{n}=\dfrac{2^n}{n}\sum_{k=0}^{n-1} \dfrac{1}{\binom{n-1}{k}}$$ This indentity proof can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?p=371496&sid=e0319f030d85bf1390a8fb335fd87c9d#p371496

also I remember this indentity stackmath have post it,But I can't find this link.

Answer 2

This is a partial answer, just underlying an interesting technique. $$\frac{a_n}{b_n}=\sum_{k=1}^n\frac{2^k}{k},\qquad (a_n,b_n)=1$$ is the opposite of the sum of the reciprocal of the roots of $$P_n(x) = \prod_{k=1}^{n}\left(x+\frac{k}{2^k}\right)=2^{-\frac{n(n+1)}{2}}\prod_{k=1}^{n}(k+2^kx),$$ so $$\frac{a_n}{b_n}=\frac{P_n'(0)}{P_n(0)}=\left.\frac{d}{dx}\left(\log P_n(x)\right)\right|_{x=0},\tag{1}$$ and we just need to estimate the $2$-adic height of $P_n(0)$ and $P_n'(0)$. The first task is quite easy: $$\nu_2(P_n(0))=-\frac{n(n+1)}{2}+\sum_{j=1}^{\infty}\left\lfloor\frac{n}{2^j}\right\rfloor\leq -\frac{n(n-1)}{2},\tag{2}$$ while for the second one some insight is needed. We need a lower bound for $\nu_2(Q_n'(0))-\nu_2(n!)$, where: $$ Q_n(x)=\prod_{k=1}^{n}(k+2^k x),$$ $$ Q_{n+1}'(0) = (n+1)\cdot Q_n'(0) + 2^{n+1}n!,$$ $$ \nu_2(Q_{n+1}'(0)) \geq \min\left(\nu_2(n+1)+\nu_2(Q_n'(0)),n+1+\sum_{j=1}^{\infty}\left\lfloor\frac{n}{2^j}\right\rfloor\right).$$ The last line trivially gives that the sequence $\delta_n=\nu_2(Q_n'(0))-\nu_2(n!)$ is non-decreasing, so, in order to prove the first point, we just need to prove the second one, that may be related to the structure of $\mathbb{Z}_{/2^n\mathbb{Z}}^*$.