设g(x)在x=0的某领域内二阶导数连续,且g(0)=1,g'(0)=2,g''(0)=1,且设

Question

关于级数的证明题
设f(x)是偶函数,在x=0的某个领域内有连续的二阶导数,且f(0)=1,f''(0)=2
证明：∑［f(1/n)-1］绝对收敛
n从1取到无穷

Answer 1

由f(x)为偶函数,且在x = 0可导,有:
f'(0) = lim{x → 0} (f(x)-f(0))/x = lim{x → 0} (f(-x)-f(0))/(-x) = lim{x → 0} (f(x)-f(-x))/(2x) = 0.
又f(x)在x = 0的某邻域内二阶连续可导,有Peano余项的Taylor展开:
f(x) = f(0)+f'(0)x+f"(0)x²/2+o(x²) = 1+x²+o(x²).
代入x = 1/n得f(1/n) = 1+1/n²+o(1/n²),即n → ∞时(f(1/n)-1)/(1/n²) = 1+o(1) → 1.
根据比较判别法,由正项级数∑1/n²收敛,可知∑(f(1/n)-1)绝对收敛.