Determine the radius of convergence for the series below. ∑ n=0
[infinity]

4(n−9)(x+9) n
Provide your answer below: R=

Answers

Answer 1

Determine the radius of convergence for the given series below:[tex]∑n=0∞4(n-9)(x+9)n[/tex] To find the radius of convergence, we will use the ratio test:[tex]limn→∞|an+1an|=limn→∞|4(n+1-9)(x+9)n+1|/|4(n-9)(x+9)n|[/tex]. The radius of convergence is 1.

We cancel 4 and (x+9)n from the numerator and denominator:[tex]limn→∞|n+1-9||xn+1||n+1||n-9||xn|[/tex]

To simplify this, we will take the limit of this expression as n approaches infinity:[tex]limn→∞|n+1-9||xn+1||n+1||n-9||xn|=|x+9|limn→∞|n+1-9||n-9|[/tex]

We can rewrite this as:[tex]|x+9|limn→∞|n+1-9||n-9|=|x+9|limn→∞|(n-8)/(n-9)|[/tex]

As n approaches infinity,[tex](n-8)/(n-9)[/tex] approaches 1.

Thus, the limit becomes:[tex]|x+9|⋅1=|x+9[/tex] |For the series to converge, we must have[tex]|x+9| < 1.[/tex]

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Related Questions

How does the number 32.4 change when you multiply it by 10 to the power of 2 ? select all that apply.
a). the digit 2 increases in value from 2 ones to 2 hundreds.
b). each place is multiplied by 1,000
c). the digit 3 shifts 2 places to the left, from the tens place to the thousands place.

Answers

The Options (a) and (c) apply to the question, i.e. the digit 2 increases in value from 2 ones to 2 hundred, and, the digit 3 shifts 2 places to the left, from the tens place to the thousands place.

32.4×10²=32.4×100=3240

Hence, digit 2 moves from one's place to a hundred's. (a) satisfied

And similarly, digit 3 moves from ten's place to thousand's place. Now, 1000=10³=10²×10.

Hence, it shifts 2 places to the left.

Therefore, (c) is satisfied.

As for (b), where the statement: Each place is multiplied by 1,000; the statement does not hold true since each digit is shifted 2 places, which indicates multiplied by 10²=100, not 1000.

Hence (a) and (c) applies to our question.

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