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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i wrote out some values and ended up with 1\[1<2x+1<1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ignore the extra 1 besides "with" lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did get 1, so wud the inequality be the final response\[1<x <0\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont think we can use the inequality as a value, so can we just stick with 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the inequality is a range of values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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)."

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0doesn't 0 work though? it might have to be (1,0]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah it does work :) i'll just write the inequality down

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 0 doesn't work because you get 3 * (1^n) which is 3 + 3 + 3 + 3... which is divergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n>\infty} 1^n = 1 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03*(lim n> inf 1^n) =3*1 = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It'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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh it's a_n = (3*(2(x) + 1)^n where we are summing up a_0 > a_inf. My mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks zaighum47 for all ur help, really apprectiate it :)
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