求数项级收敛时其和∞∑(2/(7^n)-5/(2^n) n=1

问题描述:

求数项级收敛时其和

∑(2/(7^n)-5/(2^n)
n=1

等比级数当公比小于1时收敛,且:
∑[n=1,∞] 1/7^n = (1/7 )/(1-1/7) = 1/6
∑[n=1,∞] 1/2^n = (1/2 )/(1-1/2) = 1
收敛级数的倍数,和仍收敛,且级数和为:
∑[n=1,∞] [2/7^n -5/(2^n)]
= 2*∑[n=1,∞] 1/7^n - 5*∑[n=1,∞] 1/(2^n)
= 4/3

分开成两个等比数列2/(7^n),-5/(2^n)