大家看看計算七
有錯說一下嚕~~~~
證明:\( n \in N \),\( \displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots+\frac{1}{n^2}\le 2 \)
PF:
∵\( \displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots+\frac{1}{n^2}<1+\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{(n-1) \times n}+ \)
且\( \displaystyle 1+\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{(n-1) \times n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{n-1}-\frac{1}{n}=2-\frac{1}{n}\le 2 \)
∴\( n \in N \),\( \displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots+\frac{1}{n^2}\le 2 \)