# 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 $0$.

#### 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: $2 + n + {n}^{3} = {n}^{3} \left(\frac{2}{n} ^ 3 + \frac{n}{n} ^ 3 + {n}^{3} / {n}^{3}\right) = {n}^{3} \left(\frac{2}{n} ^ 3 + \frac{1}{n} ^ 2 + 1\right)$ 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: $\sqrt{2 + {n}^{2} + {n}^{8}} = \sqrt{{n}^{8} \left(\frac{2}{n} ^ 8 + {n}^{2} / {n}^{8} + {n}^{8} / {n}^{8}\right)} = \sqrt{{n}^{8} \left(\frac{2}{n} ^ 8 + \frac{1}{n} ^ 6 + 1\right)} = {n}^{4} \sqrt{\frac{2}{n} ^ 8 + \frac{1}{n} ^ 6 + 1}$ 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
${\lim}_{n \to + \infty} {a}_{n} = {\lim}_{n \to + \infty} \frac{{n}^{3} \left(\frac{2}{n} ^ 3 + \frac{1}{n} ^ 2 + 1\right)}{{n}^{4} \sqrt{\frac{2}{n} ^ 8 + \frac{1}{n} ^ 6 + 1}} = {\lim}_{n \to + \infty} \frac{{n}^{3} \cdot 1}{{n}^{4} \cdot 1} = {\lim}_{n \to + \infty} \frac{1}{n} = 0$.

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