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SSSSS [86.1K]
3 years ago
8

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

Mathematics
1 answer:
Ugo [173]3 years ago
7 0
7. Correct

- - -

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

\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|

The limit reduces to

\displaystyle|b(x-a)|\lim_{n\to\infty}\frac{\ln n}{\ln(n+1)}=b|x-a|

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

b|x-a|

This means the radius of convergence is \dfrac1b.

- - -

10. By the ratio test, the series converges if

\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|

The limit is

\displaystyle|x^2|

and so the radius of convergence is 1.

- - -

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

\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\frac{(x-2)^n}{n^n}\right|}=\lim_{n\to\infty}\frac{|x-2|}n=0

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

- - -

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.
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Answer:

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Step-by-step explanation:

To preface, your figure is going to be a line segment, with O as your midpoint, in between points P & M.

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