Evaluate this limit:
limit {(2+ā2)n}
nāā
where, {x} = x - floor(x)
2+sqrt(2) has the conjugate of 2-sqrt(2).
One property that this pair of conjugates have is (2+sqrt(2))^n + (2-sqrt(2))^n is an integer for all integer n. You can see this by taking the binomial expansion and seeing that all the terms with radicals cancel.
Next note that 0<2-sqrt(2)<1. Then 0<(2-sqrt(2))^n<1 as well; this value also is the difference from (2+sqrt(2))^n and the first larger integer.
This is very useful to simplify our limit since we can say {(2+sqrt(2))^n} + (2-sqrt(2))^n = 1. When applied to our limit we can generate a new form:
lim (n->inf) {(2+sqrt(2))^n} = lim (n->inf) 1 - (2-sqrt(2))^n
The right limit can then be easily evaluated with basic methods with the result of 1. Therefore we have lim (n->inf) {(2+sqrt(2))^n} = 1.
Solution
