∑sin(n^2+1)π/n
当n为奇数时,
sin[(n^2+1)π/n]=sin(π + π/n) = -sin(π/n)
当n为偶数时,
sin[(n^2+1)π/n]=sin(2π + π/n) = sin(π/n)
于是原式为
-sin(π)+sin(π/2)-sin(π/3)+...
于是此数列为交错级数,又由于sin(π/n)->0,故级数收敛,再由1/n发散,知道不是绝对收敛,得证.