Question

Determine whether the sequence converges or diverges. If it converges, give the limit 11, 22, 44, 88, ...

Answer 1

they get bigger and bigger right?

Answer 2

the common ratio is r = 2... so the sequence is divergent...

Answer 3

how do i find the limit now

Answer 4

convergence the common ratio is -1 < r < 1

so if thats the case the limiting sum is

S∞=a1−r

Answer 5

\[a _{n}=11*2^{n-1}\]
\[\lim_{n \rightarrow \infty }a _{n}\rightarrow \infty and \not~\rightarrow 0\]
so series diverges

Answer 6

For sequence, actually
-1 < r ≤ 1

Answer 7

it will not converge for r=-1, because it will always alternate.
and if r=1, then the sequence (not series) converges.

Answer 8

(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)

Answer 9

@idku I'm still not understanding so the ratio for mine is 2 right?

Answer 10

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...

Answer 11

So, does your sequence tend to some (particular) value, or does the sequence not?

Answer 12

(if the sequence becomes infinitely large, that means it will diverge)

Answer 13

yeah it seems like it will go forever @idku