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
∞
∑an-k ??
n=1
Do you mean n^k an ??
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
3 answers
∞ ∞
∑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.
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.