Question

Determine the centre, radius, and interval of convergence of the power series

Answer 1

for the radius i keep getting 1

Answer 2

x<1

Answer 3

and for the interval of convergence -3<x<1

Answer 4

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(x+1)^n }{n^32^n}}\)

Answer 5

you mean that \(|x|\le1\) ?

Answer 6

the interval is right, but the solution says that the limit is 2

Answer 7

i mean the radius is 2

Answer 8

The series will converge for any x that satisfies: \(|x|\le1\)
(Because even if x=1, the alternating p-series with p=3 will converge)

Answer 9

yea thats what u got using the ratio test ,but can u look at the solution i posted

Answer 10

Wrong, not n+1, but x+1,
so you don't get what you got

Answer 11

why does it say 1/L=2

Answer 12

Look, let me do the ratio test for you,

knowing that must be:

(1) |r|<1 (and check for ±r=1)
(2) r = \(a_{n+1}/a_n\)

Answer 13

yea my ratio test simplified to |x+1|/2<1

Answer 14

x<1

Answer 15

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{(-1)^{n+1}(x+1)^{n+1}}{(n+1)^32^{n+1}}\div \frac{(-1)^{n}(x+1)^{n}}{(n)^32^{n}}\right|}\)

Answer 16

\(\large\color{black}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{(-1)^{n+1}(x+1)^{n+1}}{(n+1)^32^{n+1}}\times \frac{(n)^32^{n}}{(-1)^{n}(x+1)^{n}}\right|}\)

\(\large\color{black}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{(x+1)^{n+1}}{(n+1)^32^{n+1}}\times \frac{(n)^32^{n}}{(x+1)^{n}}\right|}\)

-1 has gone away due to absolute value.
And then algebra.

\(\large\color{black}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{n^3(x+1)}{2(n+1)^3}\right|}\)

You need this so called common ratio to be <1, so:

\(\large\color{black}{\displaystyle\left|\frac{(x+1)}{2}\right|<1}\)

\(\large\color{black}{\displaystyle\left|x+1\right|<2}\)

\(\large\color{black}{\displaystyle-2<x+1<2}\)

\(\large\color{black}{\displaystyle-3<x<1}\)

yes, andf then you check the 3 and the 1

Answer 17

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(\color{red}{-3}+1)^n }{n^32^n}}\)

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(-2)^n }{n^32^n}}\)

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 1 }{n^3}}\) ---> CONVERGES

Answer 18

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(\color{blue}{1}+1)^n }{n^32^n}}\)

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(2)^n }{n^32^n}}\)

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n }{n^3}}\) ---> CONVERGES

Answer 19

yeah but what about the radius? thats what I'm confused about

Answer 20

So interval of convergence is: x ∈ [-3,1]

And the radius is always half the interval (whether interval is closed or open)

Answer 21

all hail SolomonZelman

Answer 22

Hello, tanya!

Answer 23

but the way i learnt it using the ratio test, u get the radius by |x+1|/2<1

Answer 24

thats how it shows in my textbooks

Answer 25

Yes, and that is good, I suppose :O

Answer 26

it doesn't say anything about the radius being half of the interval

Answer 27

Just by a simple geometric series that you learn in high school.
|r|<1, and that can be easily shown (what to hear?)

Answer 28

Hiiiii, you work soo hard lad, you must take a break and drink some water, all this math

Answer 29

Was one of my favorite topics in calc 2 - power series! Juicy

Answer 30

so using |x+1|/2<1
should |x|<1

Answer 31

No, |x+1|, not |x|

Answer 32

so |x+1|<2?

Answer 33

oh, right!

Answer 34

yes i get it now! thanks a lot!

Answer 35

Yes, one more recap to finish it up....

Answer 36

also is the centre that just the half way of the interval?

Answer 37

The idea comes from:

\(\large\color{slate}{\displaystyle (1+r+r^2+r^3+r^4+r^{5})/(1-r)=1-r^5 }\)

I will show why that is so:

\(\Large \color{blue}{ \displaystyle ^{\color{red}{~~~~~~~~~~~~~~~~~~~~~~~~~{\rm r}^4~~~~~+~~~~{\rm r}^3~~~~+~~~~{\rm r}^2~~~~+~~~~{\rm r}~~~~+~~~1}}_{\Huge _\text{_______________________________}}}\)
\(\large\color{blue}{ \displaystyle -{\rm r}+1{\huge|}~~-{\rm r}^5~~+~0{\rm r}^4~+~~0{\rm r}^3~~+~~0{\rm r}^2~+~~0{\rm r}~~+~1}\)
\(\large\color{red}{ \displaystyle -{\rm r}^5~~+~~{\rm r}^4 }\)
\(\large\color{blue}{ \displaystyle ^\text{____________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^4~~+~~0{\rm r}^3 }\)
\(\large\color{red}{ \displaystyle -{\rm r}^4~~+~~{\rm r}^3 }\)
\(\large\color{blue}{ \displaystyle ^\text{_____________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^3 ~~+~~0{\rm r}^2 }\)
\(\large\color{red}{ \displaystyle -{\rm r}^3 ~~+~~~{\rm r}^2 }\)
\(\large\color{blue}{ \displaystyle ^\text{_______________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^2 ~~+~~0{\rm r} }\)
\(\large\color{red}{ \displaystyle -{\rm r}^2 ~~+~~{\rm r} }\)
\(\large\color{blue}{ \displaystyle ^\text{______________} }\)
\(\large\color{red}{ \displaystyle -{\rm r}~+~1 }\)
\(\large\color{red}{ \displaystyle -{\rm r}~+~1 }\)
\(\large\color{blue}{ \displaystyle ^\text{___________} }\)
\(\large\color{red}{ \displaystyle 0 }\)


If you agree with (and understand) the above polynomial division, then you should get an intuitive understanding of why:

\(\color{black}{ \displaystyle \color{blue}{(-{\rm r}^{\rm n}+1)}\div \color{red}{(-{\rm r}+1)} ~~= \color{green}{{\rm r}^{{\rm n}-1}~+~{\rm r}^{{\rm n}-2}~+~....~+~{\rm r}^3~+~{\rm r}^2~+~{\rm r}~+~1} }\)

\((\)For all natural number n that are greater than 1 \()\)

Thus we get:

\(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\left(r^{k-1}\right)=1+r+r^2+r^3+...+r^{n-1} = \frac{-r^n+1}{-r+1}}\)

This is where: \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\left(r^{k-1}\right)= \frac{1-r^n}{1-r}}\)

and \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\color{orangered}{a_1}\left(r^{k-1}\right)= \frac{\color{orangered}{a_1}\left(1-r^n\right)}{1-r}}\) come from.
----------------------------------
Now, convergence of an infinite geometric series will be therefore determined by the convergence of the sequence of (1-r\(^n\))/(1-r)

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}(1+r+r^2+r^3+...+r^{n-1})=\lim_{n \rightarrow ~\infty}\left(\frac{1-r^n}{1-r}\right)}\)

after applying limit properties, we get:

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(\frac{1-r^n}{1-r}\right)=\left(\frac{1}{1-r}\right)\lim_{n \rightarrow ~\infty}\left(1-r^n\right) \\[1.9 em]
\large \displaystyle =\left(\frac{1}{1-r}\right)\left(1-\lim_{n \rightarrow ~\infty}r^n\right)}\)

\({\large \displaystyle =\left(\frac{1}{1-r}\right)-\left(\frac{1}{1-r}\right)\lim_{n \rightarrow ~\infty}r^n}\)

\(\scriptsize\color{ slate }{\scriptsize{\bbox[5pt, royalblue ,border:2px solid royalblue ]{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ }}}\)

So, \(r\ne1\) (because when r=1 we get an indetermine sum for the series)
And when r>1 the limit will go into infinity.
So 0>r>1 is so far verfied.

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}(r^n)}\)

for -1<r<0, the limit will approach zero (and thus exist) as well, and therefore by a convergence of this limit for ||r|<1, we verify the convergence of the sum of the series for |r|<1.

Answer 38

Sorry for a long reply, just a geometric convergence proof, for why is it so that |r|<1.
And as you have probably heard or figured, the same idea of |r|<1 is applied here, in Ratio test (and in root test)...

EXCEPT that for r=1 the test is inconclusive which can be explained due to the fact that the limit in ratio test is not quite the common ratio.

Answer 39

wow that was a long reply, lol thanks for the effort

Answer 40

Anytime !

Answer 41

also u said that

Answer 42

shouldn't it be 2/n^3

Answer 43

i mean isnt (-2)^n(-1)^n = (2)^2n

Answer 44

No

Answer 45

Missapplication of exponent rules:

\(\LARGE a^n\cdot b^n=(a \cdot b)^n\)

Answer 46

\(\LARGE a^n\cdot b^n=(a \cdot b)^n\) YES

\(\LARGE a^n\cdot b^n=(a \cdot b)^{2n}\) NO

Answer 47

i thought u add powers in multiplication?

Answer 48

when you have the same base, you add powers:

\(\LARGE x^n\cdot x^m=x^{n+m} \) YES

Answer 49

oh, sorry my maths is really rusty

Answer 50

thanks for clearing it up

Answer 51

You just made some confusion about the rules.... anytime :0