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