## anonymous 5 years ago find all the values of x such that the series converges 3(2x+1)^n, n=0 to infinity

1. anonymous

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

2. anonymous

ignore the extra -1 besides "with" lol

3. anonymous

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.

4. anonymous

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

5. anonymous

I believe so. :)

6. anonymous

It works.

7. anonymous

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

8. anonymous

the inequality is a range of values

9. anonymous

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.

10. anonymous

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)."

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

$\lim_{n->\infty} 1^n = 1$

15. anonymous

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

16. anonymous

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.

17. anonymous

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

18. anonymous

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