How do you determine whether the sequence #a_n=sqrtn# converges, if so how do you find the limit?

Question

Cameron Alexander · Accepted Answer

#lim_(n->oo) sqrt(n) = oo#

This is quite intuitive, as when #n# grows, so does its square root, without bounds.

We can formally demonstrate it using the definition of the limit: given any #M > 0#, if we choose #N > M^2# we have for #n > N#:

#a_n = sqrt(n) > sqrt(N) > sqrt(M^2)#

so that:

#n > N => a_n > M#

which proves that:

#lim_(n->oo) sqrt(n) = oo#

Christopher Allers · Answer

undefined To determine whether the sequence $ a_n = \sqrt{n} $ converges, consider the behavior of the sequence as $ n $ approaches infinity.

If the sequence approaches a finite number $ L $ as $ n $ approaches infinity, then the sequence converges, and $ L $ is the limit of the sequence.

To find the limit of the sequence $ a_n = \sqrt{n} $, as $ n $ approaches infinity, you can use the fact that the square root function increases without bound as its input increases without bound. Therefore, the limit of $ a_n $ as $ n $ approaches infinity is infinity. Hence, the sequence $ a_n = \sqrt{n} $ diverges.

Isaiah Allen · Answer

undefined To determine whether the sequence $a_n = \sqrt{n}$ converges, you can use the properties of sequences and limits. In this case, since $a_n = \sqrt{n}$, as $n$ approaches infinity, $\sqrt{n}$ will also approach infinity. Therefore, the sequence does not converge.

If you were to find the limit of the sequence as $n$ approaches infinity, you would write it as:

$$\lim_{n \to \infty} \sqrt{n}$$

Since $\sqrt{n}$ approaches infinity as $n$ approaches infinity, the limit does not exist in the real number system. Therefore, the sequence $a_n = \sqrt{n}$ diverges.