Question

等差數(shù)列{a_n}、{b_n}的公差都不為零，若

lim

n→∞

an

bn

=3，則

lim

n→∞

b1+b2+…bn

na4n

=

1

24

．

Answer 1

分析：由條件求得 d₁=3d₂，要求的式子即 

lim

n→∞

nb1+ n(n-1) 2 •d2

n[a1 +(4n-1)•3d2]

，此式就等于n²的系數(shù)之比，運(yùn)算求得結(jié)果．

解答：解：設(shè){a_n}、{b_n}的公差分別為d₁ 和d₂，
則由

lim

n→∞

an

bn

=

lim

n→∞

a1+(n-1)d1

b1+(n-1)d2

=3，∴

d1

d2

=3，d₁=3d₂．
∴

lim

n→∞

b1+b2+…bn

na4n

=

lim

n→∞

nb1+ n(n-1) 2 •d2

n[a1 +(4n-1)•3d2]

=

lim

n→∞

b1+ n-1 2 •d2

a1+(4n-1)•3d2

 
═

lim

n→∞

b1 n-1  + d2 2

a1 n-1 +( 4n-1 n-1 ) •3d2

=

1 2 d2

12d2

=

1

24

．

點(diǎn)評(píng)：本題考查等差數(shù)列的通項(xiàng)公式，數(shù)列極限的運(yùn)算法則的應(yīng)用，屬于中檔題．