Question

Calculate the radius of convergence and the interval of convergence
(with examination of the endpoints) for the following power series:

\sum_{n=0}^\infty \frac{n-1}{n+1} x^{n}

Answer 1

Need help guys

Answer 2

the radius and interval are 2 sides of the same coin

Answer 3

find the limit as n goes to 0 of the ration an/a(n-1)

Answer 4

sorry if i do something wrong i am quite new here, i need some solution, and sorry for my english

Answer 5

the solution will come when we work thru the steps; this is a hands on study group

Answer 6

ill guide you along the way :)

Answer 7

ok thanks but i am doing it since 3 hours and i am without any solution and idea :(

Answer 8

yeah, lets start at the beginning

Answer 9

start by setting up the limit an/a(n-1)

Answer 10

\[\lim_{n\to\ inf} \frac{(n-1)\ x^{n}}{n+1}*\frac{n-1+1}{(n-1-1)\ x^{n-1}} \]
and simplify

Answer 11

cool

Answer 12

\[\lim_{n\to\ inf} \frac{(n-1)\ x}{n+1}*\frac{n}{(n-2)}\]

pull out all the stuff that doesnt have an n attached to it

\[|x|\ \lim_{n\to\ inf} \frac{(n-1)}{n+1}*\frac{n}{(n-2)}\]
\[|x|\ \lim_{n\to\ inf} \frac{n^2...}{n^2...}=|x|\]

Answer 13

so, lets set this up into your convergence:

|x| < 1 ; there is no need to simplify since its as simple as we can get, so your radius = 1 and the interval is:

-1 < x < 1

and it didnt even take me 1 hour to deal with

Answer 14

but the only way to get good at it is to get your hands dirty in here

Answer 15

:) wow, i think you are real mathematicer

Answer 16

is it end solution like that?

Answer 17

i write it all and try to understand each step

Answer 18

i figure when we get to the end, its best to stop :)

the wolf even agrees

http://www.wolframalpha.com/input/?i=sum+%28n-1%29%2F%28n%2B1%29+x%5E%28n%29

Answer 19

what that means is that the series matches the given function perfectly, it converges with the function, between -1 and 1

Answer 20

ok, is wolfram giving the hole solution? by me it doesnt do it

Answer 21

its not giving a step by step, but its affirming the outcome we reached.

|x| < 1

Answer 22

aah ok by me its doesnt do it, i am very thankful too you, its my first 10 min here and its realy cool website, and people here thank you very much

Answer 23

youre welcome, and good luck. Word of advice, there are people who want to just hand you an answer, dont trust them. Always have them verify their results :)

Answer 24

ok i will follow your advice many thanks, now i will try to understand your solution can i ask you question later if i stuck somewhere?

Answer 25

i might not be around, but there should be someone who can give you a suitable, intelligent response. just post it up on the left like you did this one. You can only ask one question at a time to limit those who are trying to use this site to cheat with. So go ahead and press the close option to ask a new question.

Answer 26

ok tthank you soo much i am very grateful.

Answer 27

Let |x|<1
\[
\sum _{n=0}^{\infty }
\frac{(n-1) x^n}{n+1}=\sum
_{n=0}^{\infty } x^n-\sum
_{n=0}^{\infty } \frac{2
x^n}{n+1}=\\
\frac 1 {1-x} -\frac 2 x \sum _{n=0}^{\infty } \frac{
x^{n+1}}{n+1}=\\
\frac 1 {1-x} -\frac 2 x (-\ln(1-x))=\\
\frac 1 {1-x}+\frac 2 x\ln(1-x) \\
\]

Answer 28

I only found the sum of your series as the difference of two very well known series. This way, I was able to write the sum as a closed form function. You do not need to do that for your problem. It is nice to know the sum if it is feasible.

Answer 29

ok thank you very much can i use your solution or above solution from amistre64 in my homework as a answer to this question? Or i need to additional calculation?

Answer 30

you should do what @amistre64 suggested. My post is useful if you were asked to find the sum of that series.

Answer 31

ok thank you very much