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

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Answer 1 · 2017-04-25T23:09:13

#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#