1)
(3^(n+2))/(5^n) =
3^2 (3/5)^n
This converges to zero.
2) cos(2/n) converges to cos(0) = 1
3)
(2^(1+3n))^1/n = 2^[1/n + 3] converges to 2^3 = 8.
I think 1) is obvious, but you still have to practice proving this rigorously. So, you have to show that for every epsilon there exists an N such that for all n>N, a_n is closer to the limiting value (in this case zero) than epsilon.
In 2) and 3) we use that if f(x) is a continuous function then you can take the limit inside the argument of f(x). So, in 2 you can take the limit of 2/n and then apply the cosine to the limit of zero.
You should prove this rule using he defintion of continuity and the definition of the limit.
For the sequences below, find if they converge or diverge. If they converge, find the limit.
an=(3^(n+2))/(5^n)
an=cos(2/n)
an=(2^(1+3n))^1/n
I am unsure about how to get started on these problems so some assistance would be great.
1 answer
1 answer