回復 1# leo790124 的帖子
當 \(n\geq2\) 時，
\(\displaystyle\frac{1}{2^{n-1}}-\frac{2^n}{\left(2^n-1\right)^2}=\frac{\left(2^n-1\right)^2-2^{n-1}\cdot2^n}{2^{n-1}\left(2^n-1\right)^2}=\frac{2^{2n-1}-2^{n+1}+1}{2^{n-1}\left(2^n-1\right)^2}=\frac{2^{n-1}\left(2^n-2^2\right)+1}{2^{n-1}\left(2^n-1\right)^2}\geq0\)
\(\displaystyle \Rightarrow \frac{2^n}{\left(2^n-1\right)^2}\leq\frac{1}{2^{n-1}}\)
故，
對任意自然數 \(n\)，恆有 \(\displaystyle T_n<\frac{2}{\left(2-1\right)^2}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots=2+\frac{\frac{1}{2}}{1-\frac{1}{2}}=3\)
