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
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 ??
∞ ∞
∑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.
To find a value of k for which the sequence an converges, we need to determine the exponent of n, denoted by k, in the expression n^k * an. If this value is finite and non-zero, then the sequence an has a finite non-zero limit.
Let's begin by expressing the formula for an in terms of n^k. First, we simplify the expression:
an = ((7n^2 + 7n + 7) / (5n^8 + 3n + 5√n))^7
Next, let's divide the numerator and denominator by n^8 to see if this helps us express the formula in terms of n^k:
an = ((7/n^6 + 7/n^7 + 7/n^8) / (5 + 3/n^7 + 5√n/n^8))^7
Now, we can see that the exponent of n in the numerator is -6, -7, and -8 respectively, and in the denominator is 0, -7, and -8 respectively. Hence, the term an can be expressed as:
an = (7n^-6 + 7n^-7 + 7n^-8) / (5 + 3n^-7 + 5√n/n^-8))^7
Comparing this expression with n^k * an, we can see that the exponent k of the n term is -6. Therefore, k = -6.
Thus, for the sequence an = ((7n^2 + 7n + 7) / (5n^8 + 3n + 5√n))^7, the value of k that makes n^k * an have a finite non-zero limit is k = -6.