Statement:
If (an) -> l, then (a1+a2+....+an/n) -> l.
Proof:
Case(i):
Let l = 0. Let, bn = a1+a2+......+an/n. Let excellon > 0 be given. Since (an) -> 0, there exist m belongs to N such that mod an < 1/2 excellon for all n>=m ------(1)
Now, Let n>=m,
mod bn = mod a1+a2+......+am+am+1+..........+an/n
<= mod a1+mod a2+ .......+mod an/n + mod am+1+...........+mod an/n
= k/n+mod am-1+.......+mod an/n [where, k=mod a1+mod a2+.......+mod an]
< k/n+(n-m/n).excellon/2 [by (1)]
< k/n+excellon/2 -----(2) [n-m/n <1]
Now, Since (k/n) -> 0, there exist n0 belongs to N such that k/n < 1/2excellon for all n>=n0 ---(3)
Let n1=max{m,n0}, then mod bn < excellon for all n>=n1. therefore, (bn) -> 0.
Case(ii):
Let, l not= 0, Since (an) -> l. (an - l) -> 0.
Therefore, ((a1 - l)+(a2 - l)+.....+(an - l)/n) -> l [by case(i)]
(a1+a2+.....an - nl/n) ->0
(a1+a2+.....an/n - l) -> 0
(a1+a2+......+an/n) -> l
Right Hand Side -> Left Hand Side. if and only if Left Hand Side -> Right Hand Side
L.H -> R.H iff. R.H -> L.H
Note:
The converce of the above theorem is not true.
Example:
Consider the (an) = (-1) power n. Then, bn = a1+a2+......+an/n
= {0 if n is ever}
{-1/n if n is odd}
Clearly, (bn) -> 0.
(an) is not convergent.
This is dummy text. It is not meant to be read. Accordingly, it is difficult to figure out when to end it. But then, this is dummy text. It is not meant to be read. Period.