Question

find all the values of x such that the series converges 3(2x+1)^n, n=0 to infinity

Answer 1

ok, i wrote out some values and ended up with -1\[-1<2x+1<1\]

Answer 2

ignore the extra -1 besides "with" lol

Answer 3

If x is greater than or equal to 1, then the function blows up at infinity and it diverges...and we know that for a geometric series (which is what we have), the absolute value of the ratio has to be less than one to converge. So, -1< (2x+1) <1 and -1 < x < 0 works.

Answer 4

i did get -1, so wud the inequality be the final response\[-1<x <0\]?

Answer 5

I believe so. :)

Answer 6

It works.

Answer 7

i dont think we can use the inequality as a value, so can we just stick with -1?

Answer 8

the inequality is a range of values

Answer 9

No, because you get 3*(-1)^n which is divergent - it just bounces up and down, 3 + 3 - 3 + 3... and it's not convergent upon any specific value.

Answer 10

If you don't want to use an inequality, you can say that "The acceptable values of x for which the series converges lie within (-1,0)."

Answer 11

doesn't 0 work though? it might have to be (-1,0]

Answer 12

yeah it does work :) i'll just write the inequality down

Answer 13

No, 0 doesn't work because you get 3 * (1^n) which is 3 + 3 + 3 + 3... which is divergent.

Answer 14

\[\lim_{n->\infty} 1^n = 1 \]

Answer 15

3*(lim n-> inf 1^n) =3*1 = 3

Answer 16

It's true that the expression for a_n converges, but when you actually compute it you get a constant stream of 3 * (1+1+1+1+1+...) which diverges.

Answer 17

oh it's a_n = (3*(2(x) + 1)^n where we are summing up a_0 -> a_inf. My mistake

Answer 18

thanks zaighum47 for all ur help, really apprectiate it :)