已知数列{xn}满足limn→∞(xn−xn−2)=0,求证:limn→∞xn−xn−1n=0.
证明:设yn=xn−xn−1,n∈N∗,则原问题转化为已知limn→∞(yn+yn−1)=0,求证:limn→∞ynn=0.
根据已知,有∀ε>0,∃N(ε)∈N∗,使得当n>N时,均有|yn+yn−1|<12ε,此时有yn=(yn+yn−1)+(−1)(yn−1+yn−2)+(−1)2(yn−2+yn−3)+⋯+(−1)n−N−1(yN+1+yN)+(−1)n−NyN,从而|yn|⩽从而\dfrac{\left|y_n\right|}{n}\leqslant \dfrac 12\varepsilon -\dfrac{N}{2n}\varepsilon+\dfrac{\left|y_N\right|}{n}<\dfrac 12\varepsilon +\dfrac{\left|y_N\right|}{n},此时只需要取N_0=\max\left\{N(\varepsilon,\left[\dfrac{2\left|y_N\right|}{\varepsilon}\right]+1\right\}即可使得\dfrac{\left|y_n\right|}{n}<\varepsilon,因此\lim_{n\to\infty}\dfrac{\left|y_n\right|}{n}=0,进而原命题得证.