Statement:
Cesaro's theorem if (an) ->a, and (bn) ->b then (a1bn+a2bn-2+anb1/n) ->ab.
proof:
Let, cn = a1bn+.....+anb1/n
Now, put an = a+rn.
so that, (rn) ->0
cn = (a+r1)bn+(a+r2)bn-1+........+(a+rn)b1/n
= a(b1+b2+....+bn)/n + r1bn+r2bn-1+.....+rnb1/n
Now, by cauchy's first limit theorem (b1+b2+.....+bn/n) ->b
Therefore, (a(b1+b2+....+bn)/n) ->ab
Hence, it is enough if be prove that (r1bn+r2bn-1+....+rnb1/n) ->0
Now, since (bn) ->b, (bn) is a bounded sequence.
Therefore, there exist a real number k > 0 such that mod bn <= k
Therefore, mod r1bn+r2bn-1+.......+rnb1/n <= k mod r1+....+rn/n
Since, (rn) ->0
Therefore, (r1+r2+......+rn/n) ->0 [by theorem cauchy's first limit]
Therefore, (r1bn+r2bn-1+........+rnb1/n) ->0
Hence the theorem.
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