anonymous
  • anonymous
Determine whether the sequence converges or diverges. If it converges, give the limit 11, 22, 44, 88, ...
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
they get bigger and bigger right?
Victoriasushchik
  • Victoriasushchik
the common ratio is r = 2... so the sequence is divergent...
anonymous
  • anonymous
how do i find the limit now

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Victoriasushchik
  • Victoriasushchik
convergence the common ratio is -1 < r < 1 so if thats the case the limiting sum is S∞=a1−r
anonymous
  • anonymous
\[a _{n}=11*2^{n-1}\] \[\lim_{n \rightarrow \infty }a _{n}\rightarrow \infty and \not~\rightarrow 0\] so series diverges
idku
  • idku
For sequence, actually -1 < r ≤ 1
idku
  • idku
it will not converge for r=-1, because it will always alternate. and if r=1, then the sequence (not series) converges.
idku
  • idku
(this is where the ratio test comes from, to show that series behaves in a way that its "r" satisfies |r|<1, but this is for series, sequence is good when r=1)
anonymous
  • anonymous
@idku I'm still not understanding so the ratio for mine is 2 right?
idku
  • idku
Look at this way (as satelite offered). YOur terms will get bigger and bigger (because you are mutliplying times 2 every time). And this will get larger and larger with no limit...
idku
  • idku
So, does your sequence tend to some (particular) value, or does the sequence not?
idku
  • idku
(if the sequence becomes infinitely large, that means it will diverge)
anonymous
  • anonymous
yeah it seems like it will go forever @idku

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