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n = 1 n 2 x n 2 n

a n = n 2 2 n so R = 2 . When x = ± 2 the series diverges by the divergence test. The interval of convergence is I = ( −2 , 2 ) .

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k = 1 k e x k e k

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k = 1 π k x k k π

a k = π k k π so R = 1 π . When x = ± 1 π the series is an absolutely convergent p -series. The interval of convergence is I = [ 1 π , 1 π ] .

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n = 1 10 n x n n !

a n = 10 n n ! , a n + 1 x a n = 10 x n + 1 0 < 1 so the series converges for all x by the ratio test and I = ( , ) .

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n = 1 ( −1 ) n x n ln ( 2 n )

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In the following exercises, find the radius of convergence of each series.

k = 1 ( k ! ) 2 x k ( 2 k ) !

a k = ( k ! ) 2 ( 2 k ) ! so a k + 1 a k = ( k + 1 ) 2 ( 2 k + 2 ) ( 2 k + 1 ) 1 4 so R = 4

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n = 1 ( 2 n ) ! x n n 2 n

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k = 1 k ! 1 · 3 · 5 ( 2 k 1 ) x k

a k = k ! 1 · 3 · 5 ( 2 k 1 ) so a k + 1 a k = k + 1 2 k + 1 1 2 so R = 2

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k = 1 2 · 4 · 6 2 k ( 2 k ) ! x k

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n = 1 x n ( 2 n n ) where ( n k ) = n ! k ! ( n k ) !

a n = 1 ( 2 n n ) so a n + 1 a n = ( ( n + 1 ) ! ) 2 ( 2 n + 2 ) ! 2 n ! ( n ! ) 2 = ( n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 1 ) 1 4 so R = 4

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n = 1 sin 2 n x n

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In the following exercises, use the ratio test to determine the radius of convergence of each series.

n = 1 ( n ! ) 3 ( 3 n ) ! x n

a n + 1 a n = ( n + 1 ) 3 ( 3 n + 3 ) ( 3 n + 2 ) ( 3 n + 1 ) 1 27 so R = 27

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n = 1 2 3 n ( n ! ) 3 ( 3 n ) ! x n

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n = 1 n ! n n x n

a n = n ! n n so a n + 1 a n = ( n + 1 ) ! n ! n n ( n + 1 ) n + 1 = ( n n + 1 ) n 1 e so R = e

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n = 1 ( 2 n ) ! n 2 n x n

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In the following exercises, given that 1 1 x = n = 0 x n with convergence in ( −1 , 1 ) , find the power series for each function with the given center a , and identify its interval of convergence.

f ( x ) = 1 x ; a = 1 ( Hint: 1 x = 1 1 ( 1 x ) )

f ( x ) = n = 0 ( 1 x ) n on I = ( 0 , 2 )

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f ( x ) = 1 1 x 2 ; a = 0

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f ( x ) = x 1 x 2 ; a = 0

n = 0 x 2 n + 1 on I = ( −1 , 1 )

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f ( x ) = 1 1 + x 2 ; a = 0

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f ( x ) = x 2 1 + x 2 ; a = 0

n = 0 ( −1 ) n x 2 n + 2 on I = ( −1 , 1 )

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f ( x ) = 1 2 x ; a = 1

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f ( x ) = 1 1 2 x ; a = 0 .

n = 0 2 n x n on ( 1 2 , 1 2 )

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f ( x ) = 1 1 4 x 2 ; a = 0

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f ( x ) = x 2 1 4 x 2 ; a = 0

n = 0 4 n x 2 n + 2 on ( 1 2 , 1 2 )

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f ( x ) = x 2 5 4 x + x 2 ; a = 2

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Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

Explain why, if | a n | 1 / n r > 0 , then | a n x n | 1 / n | x | r < 1 whenever | x | < 1 r and, therefore, the radius of convergence of n = 1 a n x n is R = 1 r .

| a n x n | 1 / n = | a n | 1 / n | x | | x | r as n and | x | r < 1 when | x | < 1 r . Therefore, n = 1 a n x n converges when | x | < 1 r by the n th root test.

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k = 1 ( k 1 2 k + 3 ) k x k

a k = ( k 1 2 k + 3 ) k so ( a k ) 1 / k 1 2 < 1 so R = 2

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k = 1 ( 2 k 2 1 k 2 + 3 ) k x k

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n = 1 a n = ( n 1 / n 1 ) n x n

a n = ( n 1 / n 1 ) n so ( a n ) 1 / n 0 so R =

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Suppose that p ( x ) = n = 0 a n x n such that a n = 0 if n is even. Explain why p ( x ) = p ( x ) .

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Suppose that p ( x ) = n = 0 a n x n such that a n = 0 if n is odd. Explain why p ( x ) = p ( x ) .

We can rewrite p ( x ) = n = 0 a 2 n + 1 x 2 n + 1 and p ( x ) = p ( x ) since x 2 n + 1 = ( x ) 2 n + 1 .

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Suppose that p ( x ) = n = 0 a n x n converges on ( −1 , 1 ] . Find the interval of convergence of p ( A x ) .

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Suppose that p ( x ) = n = 0 a n x n converges on ( −1 , 1 ] . Find the interval of convergence of p ( 2 x 1 ) .

If x [ 0 , 1 ] , then y = 2 x 1 [ −1 , 1 ] so p ( 2 x 1 ) = p ( y ) = n = 0 a n y n converges.

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In the following exercises, suppose that p ( x ) = n = 0 a n x n satisfies lim n a n + 1 a n = 1 where a n 0 for each n . State whether each series converges on the full interval ( −1 , 1 ) , or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

n = 0 a 2 n x 2 n

Converges on ( −1 , 1 ) by the ratio test

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n = 0 a 2 n x n ( H i n t : x = ± x 2 )

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n = 0 a n 2 x n 2 ( Hint: Let b k = a k if k = n 2 for some n , otherwise b k = 0 . )

Consider the series b k x k where b k = a k if k = n 2 and b k = 0 otherwise. Then b k a k and so the series converges on ( −1 , 1 ) by the comparison test.

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Suppose that p ( x ) is a polynomial of degree N . Find the radius and interval of convergence of n = 1 p ( n ) x n .

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[T] Plot the graphs of 1 1 x and of the partial sums S N = n = 0 N x n for n = 10 , 20 , 30 on the interval [ −0.99 , 0.99 ] . Comment on the approximation of 1 1 x by S N near x = −1 and near x = 1 as N increases.


This figure is the graph of y = 1/(1-x), which is an increasing curve with vertical asymptote at 1.
The approximation is more accurate near x = −1 . The partial sums follow 1 1 x more closely as N increases but are never accurate near x = 1 since the series diverges there.

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[T] Plot the graphs of ln ( 1 x ) and of the partial sums S N = n = 1 N x n n for n = 10 , 50 , 100 on the interval [ −0.99 , 0.99 ] . Comment on the behavior of the sums near x = −1 and near x = 1 as N increases.

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[T] Plot the graphs of the partial sums S n = n = 1 N x n n 2 for n = 10 , 50 , 100 on the interval [ −0.99 , 0.99 ] . Comment on the behavior of the sums near x = −1 and near x = 1 as N increases.


This figure is the graph of y = -ln(1-x) which is an increasing curve passing through the origin.
The approximation appears to stabilize quickly near both x = ± 1 .

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[T] Plot the graphs of the partial sums S N = n = 1 N sin n x n for n = 10 , 50 , 100 on the interval [ −0.99 , 0.99 ] . Comment on the behavior of the sums near x = −1 and near x = 1 as N increases.

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[T] Plot the graphs of the partial sums S N = n = 0 N ( −1 ) n x 2 n + 1 ( 2 n + 1 ) ! for n = 3 , 5 , 10 on the interval [ −2 π , 2 π ] . Comment on how these plots approximate sin x as N increases.


This figure is the graph of the partial sums of (-1)^n times x^(2n+1) divided by (2n+1)! For n=3,5,10. The curves approximate the sine curve close to the origin and then separate as the curves move away from the origin.
The polynomial curves have roots close to those of sin x up to their degree and then the polynomials diverge from sin x .

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[T] Plot the graphs of the partial sums S N = n = 0 N ( −1 ) n x 2 n ( 2 n ) ! for n = 3 , 5 , 10 on the interval [ −2 π , 2 π ] . Comment on how these plots approximate cos x as N increases.

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Practice Key Terms 3

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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