Radius of Convergence and Interval of Convergence for Given Series, Exams of Advanced Calculus

Solutions to three problems related to finding the radius of convergence and interval of convergence for given series. The series involve the limits as n approaches infinity and the comparison test and alternating series test are used to determine the convergence properties.

Typology: Exams

Pre 2010

Uploaded on 07/22/2009

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Worksheet 10/31/08
Find the radius of convergence and the interval of convergence
Problem 1. โˆž
X
n=2
(โˆ’1)nxn
4 ln(n)
Solution.
lim
nโ†’โˆž ๎˜Œ๎˜Œ๎˜Œ๎˜Œ
(โˆ’1)n+1xn+14 ln n
(โˆ’1)nxn4 ln(n+ 1)๎˜Œ๎˜Œ๎˜Œ๎˜Œ
= lim
nโ†’โˆž
|x|ln n
ln(n+ 1) =|x|
We know it to be absolutely convergent if
โ‡’ |x|<1
Checking the endpoints we see that it converges at x= 1 (alternating series test), but not
at x=โˆ’1 (comparison test).
Thus, the interval of convergence is (โˆ’1,1].
Problem 2. โˆž
X
n=1
nnxn
Solution. If x6= 0, then
lim
nโ†’โˆž
n
p|nnxn|= lim
nโ†’โˆž
nx =โˆž
Thus, this series converges only when x= 0.
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Worksheet 10/31/

Find the radius of convergence and the interval of convergence

Problem 1. โˆž โˆ‘

n=

n x

n

4 ln(n)

Solution.

lim nโ†’โˆž

n+ x

n+ 4 ln n

(โˆ’1)nxn4 ln(n + 1)

= lim nโ†’โˆž

|x| ln n

ln(n + 1)

= |x|

We know it to be absolutely convergent if

โ‡’ |x| < 1

Checking the endpoints we see that it converges at x = 1 (alternating series test), but not

at x = โˆ’1 (comparison test).

Thus, the interval of convergence is (โˆ’ 1 , 1].

Problem 2.

โˆ‘โˆž

n=

n

n x

n

Solution. If x 6 = 0, then

lim nโ†’โˆž

n

|nnxn| = lim nโ†’โˆž

nx = โˆž

Thus, this series converges only when x = 0.

Problem 3. โˆž โˆ‘

n=

(x โˆ’ 2)

n

n

n

Solution.

lim nโ†’โˆž

n

(x โˆ’ 2)n

nn

= lim nโ†’โˆž

|x โˆ’ 2 |

n

Thus the interval of convergence is (โˆ’โˆž, โˆž).