ReviewTest3_F07Solution

ReviewTest3_F07Solution - Review Problems Test 3-Solutions...

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Review Problems – Test 3-Solutions 1. Find the interval of convergence of each of the following power series: a) 1 (2 ) (1 ) 2 n n n n x n = b) 0 (3 2) 1 n n x n = + 1 1 00 ) 2 ) ) 2 2 2 lim lim lim ) 2 ) ) 2 2 ) 2 1 2 22 So the series converges if 1 1 1 2 2 2 0 4 Check endpoints: 4: 21 ( 1) ( 1) which 2 nn n n n n n n n xn x x n x nx n x n xx x + + →∞ →∞ →∞ ∞∞ == −− = +− + + < ⇒− < < ⇒− < − < ⇒ < < = −= ∑∑ is the convergent alternating harmonic series. 0: ) 1 ( 1) which diverges. 2 Interval of convergence: (0,4] n n n x = b) 1 1 ( 32 ) 1 ( ) ( ) 1 lim lim 2( ) 2 ( ) 2) 1 l i m 1 1 1 1 1 3 3 1 so the radius of convergence is 1/3. 3 Check endpoints: 11 : ( 1) which converges 31 n n n x x n n x n x n + →∞ →∞ →∞ = −+ + + =− < <− < + << < < + 1 by the Alternating Series Test. 1 1: which diverges. 1 Interval of convergence: 1 [, 1 ) 3 n x n = = + 2. Find the sum of the geometric series or show sum does not exist: a)
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1 3 6 18 54 2 ..... 4 4 16 64 63 3 42 4 3/2 3 4 6 1( 3 / 4 ) 2 1 n n ar S = ⎛⎞ =+ + + ⎜⎟ ⎝⎠ == = = b) 141 6 1 4 4 ... 1 .... so series diverges since 1 392 7 3 3 3 ++ += ++ > 3. Find the value of m for which the geometric series 23 22 2 2 ... 7 mm m +−+ converges to 5. 2 2 2 ... 2(1 ....) 7 3 9 2, 3 226 51 5 5 65 9 3 3 1 33 9 5 m m m m Sm m m m −+ + =− +− ==== + = = + + + ⇒= 4. For the series 3 1 1 n n =
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This note was uploaded on 03/31/2008 for the course MATH 2016 taught by Professor Recone during the Fall '08 term at Virginia Tech.

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ReviewTest3_F07Solution - Review Problems Test 3-Solutions...

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