Question

The series \[\large \frac{2x}{x+3}+(\frac{2x}{x+3})^{2}+(\frac{2x}{x+3})^{3}+...........\infty \] will have a non zero definite sum then find the complete x.

Answer 1

@Hero

Answer 2

ans. -1<x<3 (x not = to 0) (solution needed)

Answer 3

@lgbasallote

Answer 4

@UnkleRhaukus

Answer 5

\[\large{1+n+n^2+n^3+\dots +\infty=\frac{1}{1-n}}\]

Answer 6

ok

Answer 7

then..

Answer 8

let n = 2x/(x+3)
also note that this only holds for case 2x/(x+3) < 1

Answer 9

ok

Answer 10

using this
2x/(x+3) < 1
find the interval of convergence.

Answer 11

(-3,3)

Answer 12

@experimentX but ans.-1<x<3

Answer 13

http://www.wolframalpha.com/input/?i=abs%282x%2F%28x%2B3%29%29+%3C+1

Answer 14

\[\left| \frac{2x}{x+3} \right|<1 \Rightarrow \left| \frac{x}{x+3} \right|<\frac{1}{2} \Rightarrow \left| 1-\frac{3}{x+3} \right|<\frac{1}{2} \Rightarrow -\frac{1}{2}< 1-\frac{3}{x+3}<\frac{1}{2}\\-\frac{3}{2}< -\frac{3}{x+3}<-\frac{1}{2}\\\frac{1}{2}< \frac{3}{x+3}<\frac{3}{2}\\\frac{1}{6}< \frac{1}{x+3}<\frac{1}{2}\\ 2<x+3<6 \\ -1<x<3 \]