... an≤ a1+ an−1≤ . . . ≤ na1, and so a ≤ a1For > 0 we choose N > 0 with aN< N (a + ). For any n ≥ 1 we can writen = kN + r, where k ≥ 0 and 1 ≤ r ≤ N − 1. Thenan≤ akN+ ... sup1≤r≤Nar and we see thatlim supn→+∞ann≤ lim supk→+∞kaN+ sup1≤r≤NarkN=aNN≤ a + .This shows thatann→ a, as required.References1. P. Halmos, Measure Theory, Van Nostrand, ... Introduction to Probability and Measure Theory, Macmillan, NewDelhi, 1977.3. H. Roydon, Real Analysis, Macmillan, New York, 1968.4. G. Simmons, Introduction to Topology and Modern Analysis, McGraw-Hill,...