Asked by .
For each sequence an find a number k such that nkan
has a finite non-zero limit.
(This is of use, because by the limit comparison test the series ∑n=1∞an and ∑n=1∞n−k both converge or both diverge.)
D. a_n = ( (7n^2+7n+7)/(5n^8+3n+5√n) )^7
has a finite non-zero limit.
(This is of use, because by the limit comparison test the series ∑n=1∞an and ∑n=1∞n−k both converge or both diverge.)
D. a_n = ( (7n^2+7n+7)/(5n^8+3n+5√n) )^7
Answers
Answered by
oobleck
not sure just what you mean by
such that nkan has a finite non-zero limit.
Nowhere in your expression is there a place for k, and the series clearly converges, so maybe you can 'splain a bit.
Do you mean
∞
∑a<sub><sub>n-k</sub></sub> ??
n=1
Do you mean n^k a<sub><sub>n</sub></sub> ??
such that nkan has a finite non-zero limit.
Nowhere in your expression is there a place for k, and the series clearly converges, so maybe you can 'splain a bit.
Do you mean
∞
∑a<sub><sub>n-k</sub></sub> ??
n=1
Do you mean n^k a<sub><sub>n</sub></sub> ??
Answered by
.
∞ ∞
∑a_n and ∑n^-k are the series that converge.
n=1 n=1
I have to find k from the a_n given, and am not sure how to do so.
∑a_n and ∑n^-k are the series that converge.
n=1 n=1
I have to find k from the a_n given, and am not sure how to do so.
Answered by
oobleck
Hmmm. I'm not sure just what you are after, either.
However, as n->∞
(7n^2+7n+7)/(5n^8+3n+5√n) -> 7n^2/5n^8 -> n^-6
because all the lower powers of n don't really matter. (nor does the pesky 7/5)
Maybe that will help some.
However, as n->∞
(7n^2+7n+7)/(5n^8+3n+5√n) -> 7n^2/5n^8 -> n^-6
because all the lower powers of n don't really matter. (nor does the pesky 7/5)
Maybe that will help some.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.