Question:
Let $k$ be give the postive integers,Find the highest power of $2$ which divides the numerator of $$H_{k}=1+\dfrac{1}{3}+\dfrac{1}{5}+\cdots+\dfrac{1}{2k-1}$$
I found some case when $k=2$,then $$H_{2}=1+\dfrac{1}{3}=\dfrac{4}{3}$$,because the numerator $4$ havethe highest power of is $2$
when $k=3$,then $$H_{3}=1+\dfrac{1}{3}+\dfrac{1}{5}=\dfrac{23}{15}$$because the numerator $23$have the highest power of is $0$
when $k=4$,then $$H_{4}=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}=\dfrac{176}{105}$$because the numerator $144$ have the highest power of is $4$
So I conjecture this problem answer is $2V_{2}(k)$.then How to prove？