How prove this $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n}}{a_{n+1}}>\frac{n}{2}-\frac{1}{3}$

Question

let $$a_{n}=2^n-1$$

show that $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n}}{a_{n+1}}>\dfrac{n}{2}-\dfrac{1}{3}$$

My idea : since $$\dfrac{a_{k}}{a_{k+1}}=\dfrac{2^k-1}{2^{k+1}-1}=\dfrac{1}{2}\cdot\dfrac{2^{k+1}-1}{2^{k+1}-1}-\dfrac{1}{2}\cdot\dfrac{1}{2^{k+1}-1}=\dfrac{1}{2}-\dfrac{1}{2}\cdot\dfrac{1}{2^{k+1}-1}$$

Then I can't.Thank you

Answer 1

We have to show that

$$\sum_{n=2}^{\infty}\frac{1}{2^n-1} < \frac23$$

But this is easy, since

$$\frac13+\frac17+\frac{1}{15}+\frac{1}{31}+\cdots < \frac13\left(1+\frac12+\frac14+\frac18+\cdots\right) = \frac23$$