How to prove that $1-\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \geq \frac{2^{N-1}+1}{2^N}$?

Question

The question is in the title. Numerical computation suggests the result is true, but I don't know how to prove it rigorously.

Answer 1

I think I got it, but my answer isn't very slick. Improvements are very welcome.

Rearranging, it suffices to show that $$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \leq \frac{1}{2} - \frac{1}{2^N}.$$ Now, I can verify this inequality by hand for $N \in \{1,2\}$. Let $N > 2$. First observe that $$\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} < \frac{1}{2^{n+1}}.$$ In particular, for $n=1$, the LHS is $0$ while the RHS is $1/4$. Thus, $$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} + \frac{1}{4} < \sum_{n=1}^N\frac{1}{2^{n+1}} < \frac{1}{2},$$ which implies $$\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} < \frac{1}{4} < \frac{1}{2} - \frac{1}{2^N}$$

Answer 2

First note that $\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}=\frac{\frac{1}{2}}{2^{n-1}+1}-\frac{1}{2^{n+1}}$ (an alternative form). Also the series $\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+1}<\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}=2<\infty$ and hence it converges by the comparison test. Since $\sum_{n=1}^{N}\frac{1}{2^{n+1}}=\frac{1}{2}-(\frac{1}{2^{N+1}})$ we have $\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}=\frac{1}{2}\sum_{n=1}^{N}\frac{1}{2^{n-1}+1}-\frac{1}{2}+\frac{1}{2^{N+1}}<\frac{1}{2}\sum_{n=1}^{N}\frac{1}{2^{n-1}}-\frac{1}{2}+\frac{1}{2^{N+1}}=\frac{1}{2}(2-\frac{2}{2^{N}})-\frac{1}{2}+\frac{1}{2^{N+1}}=\frac{1}{2}-\frac{1}{2^{N}}+\frac{1}{2^{N+1}}$