Question

Question #771c5

Answer 1

The sequence diverges to infinity.

Explanation:

The terms of the sequence are all non-negative.
The ratio of consecutive terms in the sequence goes to infinity. Therefore the terms themselves go to infinity.

`(2^((n+1)^2)/((n+1)!)) /(2^(n^2)/(n!)) = 2^((n+1)^2)/((n+1)!)*(n!)/(2^(n^2))`

`= 2^((n+1)^2)/(2^(n^2))*(n!)/ ((n+1)!)`

`= 2^(n^2+2n+1)/(2^(n^2))*1/(n+1)`

`= 2^(2n+1)/(n+1)`

This sequence clearly goes to infinity.
[If one requires proof of that fact, extend
`2^(2n+1)/(n+1)` to the continuous function

`f(x) = 2^(2x + 1)/(x+1)`

and apply L'Hopital's Rule once.]

Since the terms are non-negative, and since their ratio goes to infinity as n increases, it is also true that the terms themselves go to infinity.

Still need convincing? Look at how rapidly the terms increase!

At n = 4, we have
`2^16 = 65536 and 4! = 24`
At n = 5, we have
`2^25 = 33554432 and 5! = 120`
At n = 6, the numerator is larger than 60 billion, but the denominator is just 720.