anonymous
  • anonymous
find all the values of x such that the series converges 3(2x+1)^n, n=0 to infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
ok, i wrote out some values and ended up with -1\[-1<2x+1<1\]
anonymous
  • anonymous
ignore the extra -1 besides "with" lol
anonymous
  • 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.

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anonymous
  • anonymous
i did get -1, so wud the inequality be the final response\[-1
anonymous
  • anonymous
I believe so. :)
anonymous
  • anonymous
It works.
anonymous
  • anonymous
i dont think we can use the inequality as a value, so can we just stick with -1?
anonymous
  • anonymous
the inequality is a range of values
anonymous
  • 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.
anonymous
  • 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)."
anonymous
  • anonymous
doesn't 0 work though? it might have to be (-1,0]
anonymous
  • anonymous
yeah it does work :) i'll just write the inequality down
anonymous
  • anonymous
No, 0 doesn't work because you get 3 * (1^n) which is 3 + 3 + 3 + 3... which is divergent.
anonymous
  • anonymous
\[\lim_{n->\infty} 1^n = 1 \]
anonymous
  • anonymous
3*(lim n-> inf 1^n) =3*1 = 3
anonymous
  • 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.
anonymous
  • anonymous
oh it's a_n = (3*(2(x) + 1)^n where we are summing up a_0 -> a_inf. My mistake
anonymous
  • anonymous
thanks zaighum47 for all ur help, really apprectiate it :)

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