7. Correct
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9. By the ratio test, the series will converge if
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{b^{n+1}(x-a)^{n+1}}{\ln(n+1)}}{\frac{b^n(x-a)^n}{\ln n}}\right|<1[/tex]
The limit reduces to
[tex]\displaystyle|b(x-a)|\lim_{n\to\infty}\frac{\ln n}{\ln(n+1)}=b|x-a|[/tex]
where [tex]|b|=b[/tex] because [tex]b>0[/tex] is given. So the series converges when
[tex]b|x-a|<1\implies|x-a|<\dfrac1b[/tex]
This means the radius of convergence is [tex]\dfrac1b[/tex].
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10. By the ratio test, the series converges if
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{2(n+1)}}{(n+1)(\ln(n+1))^2}}{\frac{x^{2n}}{n(\ln n)^2}}\right|<1[/tex]
The limit is
[tex]\displaystyle|x^2|<1\implies|x|<1[/tex]
and so the radius of convergence is 1.
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11. Incorrect. By the root test, the series converges for
[tex]\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\frac{(x-2)^n}{n^n}\right|}=\lim_{n\to\infty}\frac{|x-2|}n=0<1[/tex]
which means the series converges for all [tex]x[/tex], and so the interval of convergence is [tex](-\infty,\infty)[/tex].
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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.
