Question

I've solved most! Just need help with 9, 10, 14.

Would love if I could check my answers to 7, 11, 16 for which I got

7. Radius of Convergence = 1

11. Interval of Convergence = [1, 3)

16. Converges for abs value {x} < 1. Not sure how to put that into interval of convergence...plug the numbers back in?

Answer 1

7. Correct

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9. By the ratio test, the series will converge if


`\displaystyle\lim_(n\to\infty)\left|((b^(n+1)(x-a)^(n+1))/(\ln(n+1)))/((b^n(x-a)^n)/(\ln n))\right|<1`

The limit reduces to


`\displaystyle|b(x-a)|\lim_(n\to\infty)(\ln n)/(\ln(n+1))=b|x-a|`

where
`|b|=b` because
`b>0` is given. So the series converges when


`b|x-a|<1\implies|x-a|<\frac1b`

This means the radius of convergence is
`\frac1b`.

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10. By the ratio test, the series converges if


`\displaystyle\lim_(n\to\infty)\left|((x^(2(n+1)))/((n+1)(\ln(n+1))^2))/((x^(2n))/(n(\ln n)^2))\right|<1`

The limit is


`\displaystyle|x^2|<1\implies|x|<1`

and so the radius of convergence is 1.

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11. Incorrect. By the root test, the series converges for


`\displaystyle\lim_(n\to\infty)\sqrt[n]((x-2)^n)/(n^n)\right=\lim_(n\to\infty)\fracx-2n=0<1`

which means the series converges for all
`x`, and so the interval of convergence is
`(-\infty,\infty)`.

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For 14 and 16, it'll probably be too late to edit this post by the time you see this. You can try posting the remaining problems in a new question.