Question

7. Find the interval of convergence of the power series, remembering to check the endpoints if necessary: $\sum_{n=0}^{\infty} \frac{(-1)^n(x-1)^n}{2^n(n+3)}$

          7. Find the interval of convergence of the power series, remembering to check the endpoints if necessary: $\sum_{n=0}^{\infty} \frac{(-1)^n(x-1)^n}{2^n(n+3)}$
        
7. Find the interval of convergence of the power series, remembering to check the endpoints if necessary: ∑n=0^∞((-1)^n(x-1)^n)/(2^n(n+3))

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the interval of convergence of the power series, remembering to check the endpoints if necessary: 2√(n+3) n=0
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Transcript

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00:01 Hi, from the question given that consider the given sequence sum of n is equal to 0 to n is equal to 0 to infinity n times of x plus 2 the whole power n divided by 3 to the power of n plus 1.
00:15 So, here we need to find the radius of convergence and the interval of convergence.
00:19 So, let a n is equal to n by 3 to the power of n plus 1 and a n plus 1 is equal to n plus 1 divided by 3 to the power of n plus 2.
00:31 So, limit n tends to infinity a n plus 1 divided by a n is equal to n plus 1 divided by 3 to the power of n plus 2 multiplied with 3 to the power of n plus 1 divided by n.
00:48 So, that is equal to limit n tends to infinity 1 by 3 times of 1 plus 1 by n.
00:58 Now, apply limit n tends to infinity.
01:00 So, we have 1 by 3.
01:02 Therefore, we conclude that the radius of convergence radius of convergence will be 1 divided by 1 by 3 which is equal to 3 interval of convergence...
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