MA1521-S7 - NATIONAL UNIVERSITY OF SINGAPORE SEMESTER 1 2012\/2013 MA1521 Calculus for Computing Tutorial Solution 7 x1 4 1(a This is a geometric series

# MA1521-S7 - NATIONAL UNIVERSITY OF SINGAPORE SEMESTER 1...

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Unformatted text preview: NATIONAL UNIVERSITY OF SINGAPORE SEMESTER 1, 2012/2013 MA1521 Calculus for Computing Tutorial Solution 7 1. (a) This is a geometric series with common ratio x − 1 4 : ∞ ∑ n = 1 ( x − 1 ) n − 3 + ( x − 1 ) n − 1 4 n + 2 2 n − 1 = 1 6 ( x − 1 ) 2 ∞ ∑ n = 1 parenleftbigg x − 1 4 parenrightbigg n − 1 . It converges if and only if vextendsingle vextendsingle vextendsingle vextendsingle x − 1 4 vextendsingle vextendsingle vextendsingle vextendsingle < 1; that is, | x − 1 | < 4. So its radius of convergence is R = 4. (b) Let c n = n 2 n ( − 2 ) n 4 n ( 2 n + 1 ) ! . Then lim n → ∞ vextendsingle vextendsingle vextendsingle vextendsingle c n + 1 c n vextendsingle vextendsingle vextendsingle vextendsingle = lim n → ∞ parenleftbigg ( n + 1 ) 2 n + 2 2 n + 1 4 n + 1 ( 2 n + 3 ) ! · 4 n ( 2 n + 1 ) ! n 2 n 2 n parenrightbigg = lim n → ∞ ( n + 1 ) 2 2 ( 2 n + 2 )( 2 n + 3 ) parenleftbigg 1 + 1 n parenrightbigg 2 n = lim n → ∞ ( 1 + 1 n ) 2 2 ( 2 + 2 n )( 2 + 3 n ) parenleftbigg lim n → ∞ parenleftbigg 1 + 1 n parenrightbigg n parenrightbigg...
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