Question

Find the radius and interval of convergence

Answer 1

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

Answer 2

Using ratio test,

\[\lim_{n \rightarrow \infty} \left| \frac{ a_n+1 }{ a_n } \right| \implies \lim_{n \rightarrow \infty} \left| \frac{ (-1)^{n+1}(x+2)^{n+1} }{ (n+1)2^n+1 } \times \frac{ n2^n }{ (-1)^n(x+2)^n } \right|\]

Answer 3

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

Answer 4

\[\frac{ \left| x+2 \right| }{ 2 } < 1\]

Ok so if I did these steps right, I just need help with the parts after this

Answer 5

I would just solve for x, to get the interval right?

Answer 6

Yep

Answer 7

So, I'd have

\[\left| x+2 \right|<2 \implies -2 < x+2 <2 \implies -4<x<0\]

Ok so this is my interval, now how would I get the radius?

Answer 8

I guess that's the part which is troubling me, I was never sure what exactly the radius is for these questions.

Answer 9

Interval is right. The radius is simply half the length of the interval.

Answer 10

Wow really? Haha, so it's just 2?

Answer 11

Yes, that simple!

Answer 12

Bleh lol, thank you so much! :D

Answer 13

You're welcome!

Answer 14

Oh another thing, so why/ when do we have to solve for the points in the interval?

Answer 15

Like plugging in -4 for x in the series

Answer 16

Sorry for these kinds of questions, I just want to understand what exactly is going on, rather then just plugging in numbers.

Answer 17

I think the answer to **why/ when do we have to solve for the points in the interval? **
depends on the specific question.

Paul's online notes are often useful:
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx