How do you determine whether the sequence #a_n=2^n-n^2# converges, if so how do you find the limit?

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Answer 1 · 2017-01-07T19:38:33

#lim_(n->oo) 2^n-n^2 = oo#

Using the binomial expansion we have that:

#2^n = (1+1)^n = sum_(k=0)^n ((n),(k))1^n*1^n=sum_(k=0)^n ((n),(k))#

If we consider #n>3# and isolate the first three terms:

#2^n = 1+n+n(n-1) + sum_(k=3)^n ((n),(k)) = 1+n^2+sum_(k=3)^n ((n),(k))#

Or:

#2^n -n^2= 1+sum_(k=3)^n ((n),(k))#

As:

#((n),(k)) >= 1#

we have:

#sum_(k=3)^n ((n),(k)) >= sum_(k=3)^n 1 = n+1-3= n-2#

So:

#2^n -n^2 >= n-2#

and then:

#lim_(n->oo) 2^n-n^2 = oo#