问个数列相关的问题

4^k/5^(k+1)=1/5*(4/5)^k
对该式求和得Sk=1/5[4/5(1-(4/5)^k]/(1-4/5)
=1/5[4(1-(4/5)^k]
k趋于无穷时Sk趋于4/5
3^(k-1)/5^(k+1)=1/15*(3/5)^k
对(3/5)^k求和得3/5[1-(3/5)^k]/(1-3/5)
=3/2[1-(3/5)^k]
k趋于无穷时该式趋于3/2
故3^(k-1)/5^(k+1)趋于1/15*3/2=1/10
2^(k-2)/5^(k+1)=1/20*(2/5)^k
对2/5求和得2/5[1-(2/5)^k/(1-2/5)=2/3[1-(2/5)^k]
k趋于无穷时趋于2/3
故2^(k-2)/5^(k+1)趋于2/3*1/20=1/30
故原式=4/5+1/10+1/30=12/15+2/15=14/15

1/5*(4/5)^k+1/15*(3/5)^k-1/20*(2/5)^k
对此式先求和，再取极限就行了