设无穷数列的前n项和为Sn=(1⼀2+1⼀3)+(1⼀2^2+1⼀3^2)+..+(1⼀2^n+1⼀3^n),lim(n->∞)Sn=?

2024-12-14 04:09:19
推荐回答(3个)
回答1:

Sn=(1/2+1/3)+(1/2^2+1/3^2)+..+(1/2^n+1/3^n)
=(1/2+1/2^2+...+1/2^n)+(1/3+1/3^2+...+1/3^n)
=1-1/2^n+(1-1/3^n)/2
=1+1/2-1/2^n/1/(2*3^n)
所以lim(n→∞)Sn=1+1/2-0-0=1.5

回答2:

lim Sn= (1/2 +1/2^2+1/2^3+....) +(1/3+1/3^2+1/3^3 +...) ;n->∞
=(1/2)/(1- 1/2) +(1/3)/(1-1/3) =1+1/2 =1.5...ans

回答3:

分成两个等比数列