hw6 - a n is a series with positive terms. (a) Let L = max...

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Modern Analysis, Homework 6 Due July 15, 2010 1. [15] (5 pts) Rudin, Ch 3, problem 4. 2. [23] (5 pts) Rudin, Ch 3, problem 5. 3. [15] (5 pts) Rudin, Ch 3, problem 6. 4. [20] (5 pts) Rudin, Ch 3, problem 8. 5. [22] (5 pts) Let S 9 N be the set of all positive integers having no 9 digit in their decimal representations. Prove that n ∈S 9 1 n converges. Hint: For a fixed i N , how many i -digit integers in S 9 are there? What is their maximum possible reciprocal? The following problems are optional , and are worth 5 pts each: 6. [25] Fix a string X consisting of base-10 digits (for instance, “471934” is OK, “0x59F” is not). Let S X be the set of all positive integers whose base-10 representation does not contain X as a substring. Prove that n ∈S X 1 n converges. Hint: Consider base-10 n representations of integers, where n is the string length of X . 7. [25] Prove the following ratio test-like test: Suppose
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Unformatted text preview: a n is a series with positive terms. (a) Let L = max lim sup n a 2 n a n , lim sup n a 2 n +1 a n and l = min lim inf n a 2 n a n , lim inf n a 2 n +1 a n . Then 1. If L < 1 2 then a n converges. 2. If l > 1 2 then a n diverges. 3. If l 1 2 L then the test is inconclusive 1 Hint: Write m k = n a k k = n S k , with S k = a 2 k n + a 2 k n +1 + + a 2 k +1 n-1 . (b) Use this test to show that the series a n with a n = (2 n-1)!! 2 n ( n +1)! (for which both the ratio and root tests fail) converges 2 . 1 IE, there exist both convergent and divergent series for which this condition holds. 2 Traditionally, n !! does not mean ( n !)!. Rather, it usually means n ( n-2)( n-4) 1. 1...
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