Question

For what values of x does the series below converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of x does the new series converge? What is its sum?

Answer 1

\[1-\frac{ 1 }{ 2(x-3)}+\frac{ 1 }{ 4(x-3)^{2}}+...+(\frac{ -1 }{ 2 })^{n}(x-3)^{n}\]

Answer 2

are you sure this is tped correctly

Answer 3

Yup

Answer 4

What I found... The series converges for 0<x<6

Answer 5

The sum of the series within its interval of convergence is 3/x

Answer 6

(x-3)^n when all the others have it in the denominator

Answer 7

the series cant converge when x=3 since that would zero out the denominators

Answer 8

oops. I wrote the question wrong.lol
The (x-3) to the power of n are in the numerators.

Answer 9

Sorry about that.

Answer 10

a ratio test should help determine the interval of convergence

Answer 11

Yeah I did do the ratio test and got 0<x<6

Answer 12

lets dbl chk

\[\lim_{n\to inf}\frac{(-2)^n(x-3)^{n+1}}{(-2)^{n+1}(x-3)^n}\]
\[\lim_{n\to inf}\frac{x-3}{-2}\]
\[\left|\frac{x-3}{-2}\right|\lim 1\]
\[-1<\frac{x-3}{-2}<1\]
\[-2<x-3<2\]
\[1<x<5\]

Answer 13

Oh okay, that makes sense

Answer 14

but what about the abs signs?
wouldn't it be +2 in denominator?

Answer 15

oh wait nvm

Answer 16

now, when x is between 1 and 5, the geometric ratio is less than 1 right? otherwise its more than 1

Answer 17

yup! So then, the sum of the series would be 2/(x-1)?

Answer 18

when x=1 or 5, the ratio is just an alternating addition of 1s which just bounces between 0 and 1

Answer 19

well, r = (3-x)/2

for an infinite series with |r|<1

S = 1/(1-r)

1/(1-(3-x)/2)

2/(2-(3-x))

2/(2-3+x)

yes

Answer 20

Okay, I found f'(x)= n(-1/2)^n (x-3)^(n-1) is that right?

Answer 21

thats fair yes

Answer 22

The series converges -2<x<3? not sure about this one..

Answer 23

and im not sure how to find the sum for this one.

Answer 24

brb ... 5 min tops

Answer 25

lol okay :P

Answer 26

had to do stuff ... its done now lol

Answer 27

(n+1) (-1/2)^(n+1) (x-3)^((n+1)-1)
-------------------------------
n (-1/2)^n (x-3)^(n-1)



(n+1) (-1/2)
-----------
n (x-3)^(-1)

pull out all that has no n, and we are left with a lim of (n+1)/n which goes to 1

|3-x| < 2 is still our interval

Answer 28

the same one as before, 1<x<5?

Answer 29

yep

Answer 30

so the sum would be the same?

Answer 31

might want to work out a few terms, our constants are no longer constant are they

Answer 32

Alright! Thanks!

Answer 33

youre welcome

Answer 34

S=1 +2r +3r^2 +4r^3 +5r^4+...
-rS= -1r -2r^2 -3r^3 -4r^4-...
------------------------------
(1-r)S= 1 +r +r^2 +r^3 +r^4 + ...

(1-r)S = (1-r^n)/(1-r)

S = (1-r^n)/(1-r)^2 interesting

Answer 35

does it work? lol

Answer 36

lol idk.

Answer 37

1/(1+(x-3)/2)^2

when x=2, the wolf says its 4

1/(1 -1/2)^2

1/(1/2)^2 = 4

Answer 38

x=1.5 should be 16 by the wolf

1/ (1+(1.5-3)/2)^2

1/ ((2+1.5-3)/2)^2

4/ (2+1.5-3)^2

4/ (1/2)^2 = 16

Answer 39

4/(x-1)^2

x=2 gives 4

x=1.5 ... thta should be it then

Answer 40

lol, ive never tried that before ... that was fun :)

Answer 41

lol ;p glad you enjoyed that :P