Question

Results for this submission egin{tabular}{|c|c|c|c|} hline Entered & Answer Preview & Result & Correct Answer \ hline & & incorrect & frac{1}{2} \ hline & & incorrect & (-1.5, -0.5] \ hline end{tabular} 2 of the questions remain unanswered. Consider the power series [ sum_{n=1}^{infty} frac{(-2)^{n}}{sqrt{n}}(x+1)^{n}. ] Find the radius of convergence R. If it is infinite, type "infinity" or "inf". Answer: R = 1/2 What is the interval of convergence? Answer (in interval notation): Note: In order to get credit for this problem all answers must be correct.

          Results for this submission
egin{tabular}{|c|c|c|c|}
hline
Entered & Answer Preview & Result & Correct Answer \
hline
& & incorrect & frac{1}{2} \
hline
& & incorrect & (-1.5, -0.5] \
hline
end{tabular}
2 of the questions remain unanswered.
Consider the power series
[ sum_{n=1}^{infty} frac{(-2)^{n}}{sqrt{n}}(x+1)^{n}. ]
Find the radius of convergence R. If it is infinite, type "infinity" or "inf".
Answer: R = 1/2
What is the interval of convergence?
Answer (in interval notation):
Note: In order to get credit for this problem all answers must be correct.
        
Show more…
Results for this submission
egintabular|c|c|c|c|
hline
Entered     Answer Preview     Result     Correct Answer hline
        incorrect     frac12 hline
        incorrect     (-1.5, -0.5] hline
endtabular
2 of the questions remain unanswered.
Consider the power series
[ sumn=1^infty frac(-2)^nsqrtn(x+1)^n. ]
Find the radius of convergence R. If it is infinite, type "infinity" or "inf".
Answer: R = 1/2
What is the interval of convergence?
Answer (in interval notation):
Note: In order to get credit for this problem all answers must be correct.

Added by Buse Ş.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Transcript

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00:01 In this question, we are asked to find the center of the power series, the radius of convergence and the interval of convergence.
00:08 To find the center, we'll factor out 4.
00:13 We'll get negative 1 to the n times 4 to the n multiplied by x minus 1 half to the n.
00:27 Since our series is in the powers of x minus 1 half, that means that the center of the power series equals 1 half.
00:43 To find the radius of convergence, we'll use the ratio test.
00:51 We need to calculate the limit of the absolute value of a n plus 1 over a n as n goes to infinity, where a n is the general term of the series.
01:06 Since we are working with absolute values, we can ignore the negative sign.
01:12 To get a n plus 1, we simply need to replace n by n plus 1.
01:15 We'll get 4 to the n plus first power times x minus 1 half to the n plus first divided by the square root of n plus 4 multiplied by the reciprocal of a n.
01:42 We can cancel 4 to the n and we can cancel x minus 1 half to the n.
01:48 And when n goes to infinity, n plus 4 over square root of n plus 3 over the square root of n plus 4 goes to 1.
01:59 And in the limit, we'll get the absolute value of 4 multiplied by x minus 1 half.
02:09 And by the ratio test, the series converges if this limit is less than 1.
02:14 That means the absolute value of x minus 1 half must be less than 1 quarter.
02:21 And this means this immediately gives us the radius of convergence, which is 1 quarter.
02:31 Finally, let's find the interval of convergence.
02:33 We are almost there.
02:35 We just need to check the endpoints...
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