Note: in the following solution, symbols such as 1, U _ 2, …, U _ (n+ 1), U_n, 1, 2, …, n+ 1, and n is the subscript of u.

lim U _ n = A

∴ Give it to ε > 0, for ε/2 >; 0, with n, when n> is n,

| U _ n-A | & lt； ε/2

|(U_ 1+U_2+…+U_n)/n-A|

= |[U _ 1+U _ 2+…+U _ N+U _(N+ 1)+…+U _ N]/N-A |

= |[U _ 1+U _ 2+…+U _ N-NA]/N+[U _(N+ 1)-A+U _(N+2)-A+…+U _ N-A]/N |

= |[U _ 1+U _ 2+…+U _ N-NA]/N+[U _(N+ 1)-A+U _(N+2)-A+…+U _ N-A]/N |

< | U _ 1+U _ 2+…+U _ N-NA |/N+[| U _(N+ 1)-A |+| U _(N+2)-A |+…+| U _ N-A |]/N

≤| U _ 1+U _ 2+…+U _ N-NA |/N+(N-N)ε/(2n)

≤| U _ 1+U _ 2+…+U _ N-NA |/N+ε/2

Let N _ 1 = [2 | u _1+u _ 2+…+u _ n-na |/ε]+1,then n >: When n _1

|(U_ 1+U_2+…+U_n)/n-A|

< |U_ 1+U_2+…+U_N-NA|/n+ε/2

≤ε/2+ε/2

=ε

∴ lim (U_ 1+U_2+…+U_n)/n = A