5.设an=4tan"xdk, 1求∑(an+an+2) (2)对任意λ>0级数∑收敛。(9 元 解:an+an+2=A(an”x+tan"+2x) Jo(1+ tan x)tan"xdx=Jof tan"xdtanx n+1 a n十 nn+1 (1一)+ lim 223 nn十 n→对任意 级数 收敛。（ ） 求 设 (2) 0, 99 ( ); 1 (1) 5. tan , 1 2 1 4 0      +     + = n a a a n a xdx n n n n n a a x x dx n n n n (tan tan ) 2 4 2 0 + + + =  +  解 ： x xdx n (1 tan )tan 2 4 0 =  +  xd x n 4 tan tan 0 =   1 1 + = n 1 1 1 ( ) 1 1 2 1 +  + =   +  n n a a n n n ) 1 1 1 ) ( 3 1 2 1 ) ( 2 1 (1 + = − + − + + − n n sn  lim = 1. → n n s