# Does #a_n=(2+n+(n^3))/sqrt(2+(n^2)+(n^8)) #converge? If so what is the limit?

##### 1 Answer

Oct 31, 2015

The sequence converges and the limit is

#### Explanation:

When one has to work with sequences that can be reduced to a quotient of sums (and some variations of them, like in this case that embeds a square root), it's always a good idea to work as follows: at first study the numerator and find the term of the sum that grows faster and factor it out; then do the same for the denominator; finally, compare the two terms and get the behaviour of the limit.

In the specific case, we have

- Numerator:
#2+n+n^3=n^3(2/n^3+n/n^3+n^3/n^3)=n^3(2/n^3+1/n^2+1)# and we notice that the first two terms of the sum in brackets converge to#0# , so the sum in brackets converges to#1# . - Denominator:
#sqrt(2+n^2+n^8)=sqrt(n^8(2/n^8+n^2/n^8+n^8/n^8))=sqrt(n^8(2/n^8+1/n^6+1))=n^4 sqrt(2/n^8+1/n^6+1)# and we notice that the first two terms of the sum under the square root converge to#0# , so the square root converges to#1# .

So the limit becomes

This shows that the sequence converges and the limit is