[math]B_n=2B_n-B_n[/math]
[math]B_n=\frac{1}{2^n-1}-\frac{1}{2^{n+1}-2}[/math]
[math]S_n=\sum^n_{i=1}({\frac{1}{2^i-1}-\frac{1}{2^{i+1}-2}})[/math]
[math]S_n=\frac{1}{2^1-1}-\frac{1}{2^2-2}+\frac{1}{2^2-1}-\frac{1}{2^3-2}+\cdots+\frac{1}{2^n-1}-\frac{1}{2^{n+1}-2}[/math]
[math]S_n=\frac{1}{2^1-1}-\frac{1}{(2^2-1)(2^2-2)}-\cdots-\frac{1}{(2^n-1)(2^n-2)}-\frac{1}{2^{n+1}-2}[/math]
[math]S_n<\frac{1}{2^1-1}-\frac{1}{(2^2-1)(2^2-2)}=1-\frac16=\frac56[/math]
原式得证.
