How to prove that $1-\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \geq \frac{2^{N-1}+1}{2^N}$? The question is in the title. Numerical computation suggests the result is true, but I don't know how to prove it rigorously.

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 $\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}=\frac{1}{2}\sum_{n=1}^{N}\frac{1}{2^{n-1}+1}-\frac{1}{2}+\frac{1}{2^{N+1}}<\frac{1}{2}\sum_{n=1}^{N}\frac{1}{2^{n-1}}-\frac{1}{2}+\frac{1}{2^{N+1}}=\frac{1}{2}(2-\frac{2}{2^{N}})-\frac{1}{2}+\frac{1}{2^{N+1}}=\frac{1}{2}-\frac{1}{2^{N}}+\frac{1}{2^{N+1}}$
