Notice there was no need to use the absolute

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². Notice: There was no need to use the absolute convergence of the series n =1 b n for this part. This is known as Mertens’ Theorem: If a n converges absolutely, b n converges , then c n converges and equals a n · b n . 2. Problem 20 of Chapter VII of Rosenlicht (pp. 162-63). Solution. The radii are 1 , 1 , 1 , , 1 /e . 3. Problem 22 of Chapter VII of Rosenlicht (p. 163). Solution. Let f ( z ) = n =0 c n ( z - a ) n . The assumption is that not all coefficients are zero; we have to prove there exists δ > 0 such that f ( z ) 6 = 0 for 0 < | z - a | < δ . The exercise is stated for the real case, but the same result holds in the complex case. Let n be the first index with c n 6 = 0; that is n N 0 is such that c k = 0 if k < n , c n 6 = 0. Then f ( z ) = c n ( z - a ) n + c n +1 ( z - a ) n +1 + · · · = ( z - a ) n g ( z ) where g ( z ) = X k = n c k ( z - a ) k - n = X k =0 c k + n ( z - a ) k . It is easy to see that the radius of convergence of the series defining g is the same as that of f . So g is a continuous function in | z - a | < r and g ( a ) = c n 6 = 0. By continuity, there is δ > 0 such that g ( z ) 6 = 0 if | z - a | < δ . Since | z - a | n 6 = 0 if z 6 = a , it follows that f ( z ) 6 = 0 for 0 < | z - a | < δ . 4. C ( K ) is a Banach space. Solution. The Theorem on page 90 of Rosenlicht has a much more general result of which the completeness of C ( K ) is an immediate conse- quence.
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