Math2400
  • Math2400
Match each of the power series with its interval of convergence.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Math2400
  • Math2400
Is this right? i only get one try so I wanted to be sure i got them ><
anonymous
  • anonymous
Ill work them myself real quick to see
Math2400
  • Math2400
aww thanks @Concentrationalizing i can post my work if that's easier

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I have better luck working it myself. Ive noticed that I can miss something a student does wrong if I just scan their work.
Math2400
  • Math2400
haha sounds good :)
anonymous
  • anonymous
1st one is fine.
Math2400
  • Math2400
kk :)
anonymous
  • anonymous
3 and 4 need to be flip-flopped
anonymous
  • anonymous
Ill show the work
Math2400
  • Math2400
kk thanks :)
Math2400
  • Math2400
so it is C,B,D,A
Math2400
  • Math2400
And if i could see ur work i'd appreciate that :)
anonymous
  • anonymous
Via root test on #3 \[\lim_{n \rightarrow \infty} \sqrt[n]{\left| \frac{ (x-9)^{n} }{ 9^{n} } \right|}\] \[= \left| x-9 \right|\cdot \lim_{n \rightarrow \infty}\frac{ 1 }{ 9 } = \frac{ 1 }{ 9 }\left| x-9 \right|\] All the n's cancel and this is your limit. The conditions for root test are the same as for ratio test, we need to be less than 1. Thus we have: \[\frac{ 1 }{ 9 }\left| x-9 \right| < 1 \implies \left| x-9 \right| < 9\] which would give you the (0,18) result (since its multiple choice, I assume we dont need to actually check the endpoints)
anonymous
  • anonymous
i will not butt in and let @Concentrationalizing finish, but i am willing to bet you can guess at least two of these doing no work now i will go away
anonymous
  • anonymous
Via ratio test on #4 \[\lim_{n \rightarrow \infty}\left| \frac{ (x-9)^{n+1} }{ (n+1)!9^{n+1} }\cdot \frac{ 9^{n}n! }{ (x-9)^{n} } \right|\] \[= \left| x-9 \right|\lim_{n \rightarrow\infty}\left| \frac{ 9^{n}n! }{ 9(n+1)9^{n}n! } \right|\] \[= \left| x-9 \right|\lim_{n \rightarrow \infty}\left| \frac{ 1 }{ 9(n+1) } \right| = 0\] So all values of x work since we got a result of 0
anonymous
  • anonymous
I could guess them, but I'm not comfortable enough doing that yet, I'd rather just do the work and make sure I'm correct, lol @satellite73 Anyway, normally these aren't multiple choice x_x
Math2400
  • Math2400
haha thanks :)
anonymous
  • anonymous
No problem :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.