Question

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

Answer 1

The sequence converges and the limit is `0`.

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
`lim_{n to +infty} a_n=lim_{n to +infty} [n^3(2/n^3+1/n^2+1)]/[n^4 sqrt(2/n^8+1/n^6+1)]=lim_{n to +infty} [n^3 * 1]/[n^4 * 1]=lim_{n to +infty} 1/n=0`.

This shows that the sequence converges and the limit is `0`.