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Mathematics 10 Online
OpenStudy (anonymous):

determining convergent or divergent sequence

OpenStudy (anonymous):

\[a_{n}= \frac{ 3^{n+1} }{ 5^{n} }\]

OpenStudy (anonymous):

I would say divergent but I am not sure how to show it in a mathematical way

OpenStudy (xapproachesinfinity):

how do you determine if a sequence is div or cvg?

OpenStudy (anonymous):

taking the limit

OpenStudy (xapproachesinfinity):

well take the limit then, what are waiting for?

OpenStudy (xapproachesinfinity):

you*

OpenStudy (anonymous):

so that's all I would say\[\lim_{n \rightarrow \infty}\frac{ 3^{n+1} }{ 5^{n} }= \infty\]

OpenStudy (xapproachesinfinity):

how is that you found infinity?

OpenStudy (xapproachesinfinity):

you have infinity over infinity which is indeterminate remember!

OpenStudy (anonymous):

but using l'hopital would just continue to yield an indeterminate

OpenStudy (xapproachesinfinity):

analytically the bottom goes to infinity really fast than the top when try plugging some values you find the bottom is always in rush than the top so that would mean the limit is zero

OpenStudy (xapproachesinfinity):

no using l'hopital would make more apparent that it is going to zero

OpenStudy (xapproachesinfinity):

note that \[\lim_{n\to \infty}r^n=0\] exactly when \[|r|<1\]

OpenStudy (xapproachesinfinity):

so the sequence here cvg yes?

OpenStudy (xapproachesinfinity):

you could rewrite it as \[\lim_{n\to \infty}3\left(\frac{3}{5}\right)^n\]

OpenStudy (xapproachesinfinity):

3/5<1 so the the limit is zero

OpenStudy (anonymous):

ok that makes it much clearer

OpenStudy (anonymous):

can you explain to me the r^n statement you had from earlier though that was not covered during class for sequences but it was for series

OpenStudy (xapproachesinfinity):

remember you exponential when r<1 what is does it goes to zero ( i mean the graph when we dealt with functions) same applies to sequences

OpenStudy (xapproachesinfinity):

r=3/5 is just the ratio of the sequence note that this sequence is geometric and we usually denote the ratio with an r

OpenStudy (xapproachesinfinity):

i should of explained this way \[a^x,~~ when ~~|a|<1\] the lim goes to zero when x grows large

OpenStudy (xapproachesinfinity):

recall this from calc1 and even precal :)

OpenStudy (anonymous):

that makes sense thank you

OpenStudy (xapproachesinfinity):

no problem

OpenStudy (xapproachesinfinity):

another way is to show that \[a_n\] is monotonic and bounded this one seems to be bounded for sure and even monotonic I'm saying because we know the behavior of exp function so we embed that to sequence

OpenStudy (xapproachesinfinity):

if you are confused by monotonic stuff don't worry it is just a fancy way of saying it is strictly increasing or strictly decreasing :)

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