Question

Please Help! Find the interval of convergence of the power series...

Answer 1

\[\sum_{n=1}^{\infty} ((2x+3)^n)/n\]

Answer 2

ratio test for this one

Answer 3

\[\frac{(2x+3)^{n+1}n}{(n+1)(2x+3)^n}=\frac{(2x+3)n}{n+1}\]
\[\lim_{n\to \infty}\frac{n}{n+1}=1\] so you have to make sure that
\[|2x+3|<1\]

Answer 4

easy enough to solve, and with a little practice you can find what you need to solve with your eyeballs

Answer 5

i don't mean the problem is easy enough, i mean solving
\[|2x+3|<1\] is easy enough

Answer 6

k?

Answer 7

Yup got that

Answer 8

just to make sure, is the interval (-2, infinity) ?

Answer 9

oh heck no!!

Answer 10

I mean -1

Answer 11

imagine what you would get if you put \(x=0\) you would get
\[\sum\frac{3^n}{n}\] no chance of that converging

Answer 12

you have to make sure that
\[|2x+3|<1\] right?

Answer 13

Yup

Answer 14

i.e. the numerator has to be small, so you can raise it to large powers and get smaller numbers

Answer 15

i will write the algebra
\[|2x+3|<1\]
\[-1<2x+3<1\]
\[-4<2x<-2\]
\[-2<x<1\]

Answer 16

all this fancy laurant series stuff, don't forget the basics!

Answer 17

Ohh that was simple lol I was complicating it! Thanks a lot!!

Answer 18

hold on i forgot the basics too!

Answer 19

\[-4<2x<-2\]
\[-2<x<-1\] is what i was after

Answer 20

Haha welcome to the club

Answer 21

hey i started the club