Question

conditionally convergent ?

Answer 1

\[\sum_{1}^{\infty}4^n x^{2n}/n\]

Answer 2

it is conditionally convergent at x=-1/2 WHY

Answer 3

it is conditionally convergent at x=-1/2 WHY

Answer 4

tommorrow i have exam about them

Answer 5

it is conditionally convergent at x=-1/2 WHY

Answer 6

the interval of convergence is [-1/2,1/2)

Answer 7

it is conditionally convergent at x=-1/2 WHY

Answer 8

it conditionally divergent at -1/2

Answer 9

it is conditionally convergent at x=-1/2 WHY

Answer 10

but why at 1/2 it coverges again am i wrong?

Answer 11

Wolfram says sum is -log(1-4x^2) so

when 4x^2 = 1 ie x is plus minus 1/2 u have log 0

Answer 12

Tuna, do the ratio test:
\[\lim_{n \rightarrow \infty} \left| \frac{4^{n+1}x^{2(n+1)}}{n+1}*\frac{n}{4^n x^{2n}} \right|=\lim_{n \rightarrow \infty} \left| \frac{4 x^2 n}{n+1} \right|=\lim_{n \rightarrow \infty} \left| \frac{4x^2}{1+\frac{1}{n}} \right|=4x^2>1\]
\[4x^2>1 \implies x^2>\frac{1}{4} \implies x> \pm \frac{1}{2}\]
So your interval of convergence is: (-1/2,1/2)
Plug in the end points to see if those series converge, if they do they are included. In this case, the -1/2 series CONVERGES and the 1/2 DIVERGES.
So your interval is:
\[[\frac{-1}{2},\frac{1}{2})\]

Answer 13

thank you Malevolence !

Answer 14

No problem :P

Answer 15

but when i i insert-1/2 it becomes divergent ?

Answer 16

Plug in -(1/2) in for x.
Giving:
\[\sum_{n=1}^{\infty}\frac{4^n (\frac{-1}{2})^{2n}}{n}=\sum_{n=1}^{\infty}\frac{4^n (\frac{1}{4})^n}{n}=\sum_{n=1}^{\infty}\frac{1}{n} \rightarrow \infty\]
Hmmmmmmmmmmmm....I see what you're saying... I have no idea then :shrug:

Answer 17

ok smt wrong with my book ))

Answer 18

Then if you put in +1/2 I hope you see it diverges as well.

Answer 19

yes

Answer 20

Okay :) So, unless my thinking is wrong, the interval should be:
\[(\frac{-1}{2},\frac{1}{2})\]

Answer 21

i agree but the book doesnt agree with us Thomas's Calculus . we are right