Question

Find the radius of convergence and interval of convergence of the series. See first post for question. Second post for work I have so far.

Answer 1

\[\sum_{n=0}^{\infty}(-1)^{n}\frac{ (x-2)^{n} }{ n ^{2}+1 }\]

Answer 2

...and the process of course is to find the limit of the ratio of (n+1)/n

Answer 3

\[\Huge \lim \frac{(-1)^{n+1}\frac{ (x-2)^{n+1} }{ (n+1) ^{2}+1 }}{(-1)^{n}\frac{ (x-2)^{n} }{ n ^{2}+1 }}\]


\[\Huge\lim\frac{(-1)\frac{ (x-2) }{ (n+1) ^{2}+1 }}
{\frac{1}{ n ^{2}+1 }}\]

\[\Huge|x-2|~\lim\frac{{ n ^{2}+1 } }{ (n+1) ^{2}+1 }\]

\[\Huge|x-2|*1\]

Answer 4

|x-2| < 1

-1 < x-2 < 1

1 < x < 3

Answer 5

\[\large \lim_{n \rightarrow \infty}\left| \frac{ (-1)^{n+1}(x-2)^{n+1} }{ (n+1)^{2}+1 }*\frac{ n ^{2}+1 }{ (-1)^{n}(x-2)^{n} } \right|\]

\[\large \left| -1(x-2) \right| \lim_{n \rightarrow \infty}\left| \frac{ n ^{2}+1 }{n ^{2}+2n+2 } \right|\]

This is what I had scribbled out not sure if it is even the right path.

Answer 6

ah ok so i was and i had the 1<x<3 on the side ok.

so is that the answer then?

Answer 7

the interval of convergence is 1 to 3, the radius is just the distance from 2 to 3 ...

Answer 8

if we had come up with a fraction as a limit, then the radius would have been something spectacular to behold ... but as is, its just 1 lol

Answer 9

How do i know it is from 2 to 3 then? Sorry this teacher barely speaks English and the book I have not been able to understand.

Answer 10

the end of the process: |x-2| defines an expression that has to be < 1 to relate to a geometric series that is convergent

|x-2| < 1 simply tells us that (x-2) is between -1 and 1, subtract off the -2 to define the interval value of x

Answer 11

-1 < x-2 < 1
+2 +2 +2
-------------
1 < x+0 < 3

Answer 12

Ok so that is the interval which part shows the radius?

Answer 13

the radius ... think of a circle:

Answer 14

if we center a circle between 1 and 3, what is the length of its radius?

Answer 15

ok so the interval of convergence is similar/or equal to diameter thank you

Answer 16

interval of convergence is "related" to, not equal to :) but yes