Hey guys
Can you check my proof?
If a_n -> L and b_n = (a1 + .. + a_n)/n, show that b_n -> L.
pf) For any e>0, there is natural number N s.t. |a_n - L | < e/2(n-N) when n>N.
Fix such N
By archimedian property, there is natural number n s.t. |a1-L| + ... + |a_N - L | < en/2
For N ,
|b_n - L| = |a1 - L + .... + aN - L + a_N+1 - L + ... + a_n -L | / n < |a1 - L|/n + ... + |aN - L|/n + |a_N+1 - L|/n + ... + |a_n-L|/n
< e/2 + (n-N) * e/2(n-N) = e.
Hence b_n -> L