Question

(b) \sum_{n=0}^{\infty} \frac{(10)^n x^n}{n!} (c) \sum_{n=0}^{\infty} \frac{(-1)^n (x+1)^n}{\sqrt{n+2}} (d) \sum_{n=1}^{\infty} \frac{(x-2)^n}{n^3}

          (b) \sum_{n=0}^{\infty} \frac{(10)^n x^n}{n!}
(c) \sum_{n=0}^{\infty} \frac{(-1)^n (x+1)^n}{\sqrt{n+2}}
(d) \sum_{n=1}^{\infty} \frac{(x-2)^n}{n^3}
        
(b) ∑n=0^∞ ((10)^n x^n)/(n!)
(c) ∑n=0^∞ ((-1)^n (x+1)^n)/(√(n+2))
(d) ∑n=1^∞ ((x-2)^n)/(n^3)

Added by Karen H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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=0 n! n=0 n+2 (d) n=i n3"
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Transcript

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00:01 For this problem here to determine the radius and interval of convergence, for the series summation from n equals 0 to infinity, of x raise a power of n over, n plus 1, raise to n times 2, raise a power of n plus 1.
00:15 Now this is equal to the summation from n equals 0 to infinity of x raise to n over, n plus 1 raise a n times 2 times 2 raise a power of n.
00:27 So for the radius and interval of convergence, we will use root test.
00:33 Now let's say a sub n is equal to x raise to n over n plus 1 raise a n times 2 times 2 raise a power of n.
00:43 Then in root test, i'm going to take the limit as an approach as infinity of the absolute value of the nth root of a sub n.
00:54 So that's limit as n approaches infinity, the absolute value of the nth root of x raise a n over, n plus 1 raise a n times 2 times 2 raise a power of n...
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