Question

find the interval of convergence of the following power series :

[1*3*5*7*...*(2n+1)/4*9*14*...*(5n-1)] * (x-1)^n

find the interval of convergence of the following power series :

[1*3*5*7*...*(2n+1)/4*9*14*...*(5n-1)] * (x-1)^n

@Mathematics

Answer 1

This is a good question

Answer 2

let me write it again first

Answer 3

Infinite Series :O .. OMG !!
I haven't studied this stuff yet.

Answer 4

\[\sum_{n = 0}^{\infty}\frac{1*3*5*7*...*(2n+1)}{4*9*14*...*(5n-1)}(x-1)^n\]

Answer 5

have you already found the radius of convergence?

Answer 6

we can simplify the denominator a bit.

Answer 7

Nope..to find the radius and interval of convergence i need the general term of the series

Answer 8

the denominator is actually 2^n mutiplied by the sequence of odd numbers

Answer 9

that way you can actually simplify the whole thing to
\[\sum_{n=0}^{\infty}{\frac{1}{2^n}}(x-1)^n\]

Answer 10

now we can use the root test!

Answer 11

woah..awesome..thanks alot. but can you tell me how the denominator is 2^n * odd series? i dont see it :S

Answer 12

yeah you're right.... I don't think it is 2^n * odd series

Answer 13

hmm..cant figure it out

Answer 14

lets just use the ratio test right away

\[\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}| = \lim_{n\rightarrow\infty} \frac{2n+3} {5n+4}*(x-1)\implies|x-1|<\frac{5}{2}\]

Answer 15

but if we use the ratio test right away it only takes care of one term right? thought we need a general term..

Answer 16

so our interval of convergence is \[-\frac{3}{2}<x<\frac{7}{2}\]

Answer 17

nah we can divide all the terms out.

Answer 18

I think we must test those limits now :-D

Answer 19

lets try the left one, -3/2

this makes it \[\sum_{n = 0}^{\infty}\frac{1*3*5*7*...*(2n+1)}{4*9*14*...*(5n-1)}(-\frac{5}{2})^n\]

Answer 20

hmm okay yeah now if its convergent at both the end points of the interval then its absolutely convergent, and if not it'll be conditionally covergent at one of them which is what i needed. thanks alot!

Answer 21

let's look for a pattern.

Answer 22

\[\sum_{n = 0}^{\infty}\frac{1*3*5*7*...*(2n+1)(5^n)}{4*9*14*...*(5n-1)(2^n)}(-1)^n\]=\[\sum_{n = 0}^{\infty}\frac{5*15*25*35*...*5(2n+1)}{8*18*28*...*2(5n-1)}(-1)^n\]

Answer 23

be right back..