determining convergent or divergent sequence
\[a_{n}= \frac{ 3^{n+1} }{ 5^{n} }\]
I would say divergent but I am not sure how to show it in a mathematical way
how do you determine if a sequence is div or cvg?
taking the limit
well take the limit then, what are waiting for?
you*
so that's all I would say\[\lim_{n \rightarrow \infty}\frac{ 3^{n+1} }{ 5^{n} }= \infty\]
how is that you found infinity?
you have infinity over infinity which is indeterminate remember!
but using l'hopital would just continue to yield an indeterminate
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
no using l'hopital would make more apparent that it is going to zero
note that \[\lim_{n\to \infty}r^n=0\] exactly when \[|r|<1\]
so the sequence here cvg yes?
you could rewrite it as \[\lim_{n\to \infty}3\left(\frac{3}{5}\right)^n\]
3/5<1 so the the limit is zero
ok that makes it much clearer
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
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
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
i should of explained this way \[a^x,~~ when ~~|a|<1\] the lim goes to zero when x grows large
recall this from calc1 and even precal :)
that makes sense thank you
no problem
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
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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